# CBSE Class 12 Maths Revision Notes Chapter 2

## Class 12 Mathematics Chapter 2 Notes

Mathematics is a subject that demands lots of practice. Students need to have an in-depth knowledge of all basic concepts to understand more complex topics of the subject. The Class 12 Mathematics Chapter 2 notes introduce various concepts of the inverse of trigonometric functions. This chapter is very important and has high weightage in the board examinations. These concepts are also to be used to solve MCQs asked in JEE Mains and other national-level competitive examinations.

Extramarks provides stepwise and detailed Class 12 Mathematics Chapter 2 notes, keeping a note on its overall importance. It ensures sufficient practice through problem-solving and helps to clear doubts and queries.

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## Key Topics Covered In Class 12 Mathematics Chapter 2 Notes

The key topics covered in class 12 Mathematics chapter 2 notes include the following.

### Inverse function:

Let x and y be two functions such that y = f(x) and x= g(y). Composition of function fog(y) = f(g(y)) and gof(y) = g(f(y))=x then f and y are invertible or inverse of each other, i.e., g = f-1. We can say that, if y=f(x) then x = f-1(y).

### Inverse trigonometric function:

If y = f(x) = sin x then its inverse x = sin-1(y), where y ∈ [2 , 2 ] and x ∈ [-1,1]. This is applicable for all trigonometric functions.

Domain, Range and Graphs of all Inverse trigonometric functions:

 Functions Domain Range x= sin (y)  y = sin-1(x) [-1, 1] [–2 , 2 ] x= cos (y) y = cos-1(x) [-1, 1] [0, ] x= tan (y) y = tan-1(x) [∞, -∞] or R [–2 , 2 ] x= cot (y) y = cot-1(x) [∞, -∞] or R [0, ] x= sec (y) y = sec-1(x) R – [-1, 1] [0, ] – {2} x= cosec (y) y = cosec-1(x) R – [-1, 1] [–2 , 2 ] – {0}

### Graphs of Inverse Trigonometric functions:

1.  y= sin-1(x)

2. y= cos-1(x)

1. y = tan-1(x)

1. y= cot-1(x)

1. y = cosec-1(x)

1. y= sec-1(x)

NOTE:

1. sin-1(x) and tan-1(x) are increasing functions, whereas cos-1(x) and cot-1(x) are decreasing functions over their domain.
2. sin-1(x)  and (sin (x))-1 are different and should not be confused.

### Properties:

Property 1: Relation between two trigonometric functions.

1. sin -1(1x)= cosec-1(x) values of x ∈(−∞,1] ∪ [1,∞)
2. cos -1(1x)= sec-1(x) values of x ∈ (−∞,1] ∪ [1,∞)
3. tan -1(1x)= cot-1(x) values of x > 0

= – + cot-1(x) values of x < 0

Property 2: Negative angle Inverse Trigonometric identities

1. sin -1(x)= -sin-1(x) values of x ∈ [-1, 1]
2. tan -1(x)= -tan-1(x) values of x ∈ R
3. cosec -1(x)= -cosec-1(x) values of x ∈ (−∞,−1]∪[1,∞)
4. cos -1(x)= -cos-1(x) values of x ∈ [-1, 1]
5. sec -1(x)= -sec-1(x) values of x ∈ (−∞,−1]∪[1,∞)
6. cot -1(x)= -cot-1(x) values of x ∈ R

Property 3:

1. sin -1(x)+ cos -1(x)= 2 values of x ∈ [-1, 1]
2. tan -1(x)+ cot -1(x)= 2 values of x ∈ R
3. sec -1(x)+ cosec -1(x)= 2 values of x ∈ (−∞,−1]∪[1,∞)

Property 4

1. tan -1(x) + tan -1(y)= tan -1x + y1-xy values of xy < 1
2. tan -1(x)tan -1(y)= tan -1x – y1+xy values of xy > -1
3. tan -1(x)tan -1(y)= + tan -1(x + y1-xy) values of xy > 1; x,y= 0

Property 5:

1. 2tan -1(x)= sin -1(2x1+x2), x 1
2. 2tan -1(x)= cos -1(1-x21+x2), x 0
3. 2tan -1(x)= tan -1(2x1-x2), -1x 1

Property 6:

1. sin(sin -1(x)) = x, 2 x2

= -x, 2 x32

1. cos(cos -1(x)) = x, 0 x

= 2-x, x 2

1. tan(tan -1(x)) = –-x, x [32 ,- 2

= x, x [2 , 2 ]

= x – , x [2 ,32 ]

= x – 2, x [32 ,52 ]

1. sin -1(x) + sin -1(y) = sin -1(x1- y2 +y1- x2   )
2. sin -1(x) sin -1(y) = sin -1(x1- y2 -y1- x2   )
3. cos -1(x) + cos -1(y) = cos -1(xy –1- y2 1- x2   )
4. cos -1(x) cos -1(y) = cos -1(xy +1- y2 1- x2   )
5. tan -1(x) + tan -1(y) + tan -1(z)= tan -1x + y + z – xyz1 – xy – yz – xz , if x, y, z>0 & xy+ yz+ zx<1
6. If tan -1(x) + tan -1(y) + tan -1(z)= , then x+ y+ z= xyz
7. If tan -1(x) + tan -1(y) + tan -1(z)= 2, then xy+ yz+ z
8. sin -1(x) + sin -1(y)+ sin -1(z)= 32, x= y= z= 1
9. cos -1(x) + cos -1(y)+ cos -1(z)= 3, x= y= z= -1

Remember:

• When approaching the equations, square them so that it becomes easier to solve. Sometimes the answers or roots of the equation will not satisfy the original equation.
• Do not cancel common factors involving unknown angles on the left-hand side (LHS) and righthand side (RHS) of an equation because it may be the solution of the given equation.
• The solution of any equation, including sec θ or tan θ, can never be in the form (2n + 1) π / 2.
• The solution of any equation, including cosec θ or cot θ, can never be in the form θ = nπ.

## Class 12 Mathematics Chapter 2 Notes Exercises & Answer Solutions.

Chapter 2 of mathematics class 12 helps students to learn the inverse trigonometric functions, their range domain and graphs. It is an important chapter and holds relevance in the 12th board as well as in various entrance exams like JEE, State Engineering Entrance Exams, etc. Extramarks has prepared class 12 mathematics chapter 2 notes which include all Exercise and Answer Solutions. Students can refer to the solutions provided for exercise questions and seek help if they face any difficulties or doubts. The CBSE revision notes include detailed information about theorems, formulas, and derivations in a very descriptive way.

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Q.1

$\mathrm{If}{\mathrm{sin}}^{-1}\left(\frac{3}{5}\right)\mathrm{}=\mathrm{x},\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{cos}\mathrm{x}.$

Ans

$\begin{array}{l}\mathrm{Here},{\mathrm{sin}}^{-1}\frac{3}{5}\mathrm{}=\mathrm{}\mathrm{x}\\ ⇒\mathrm{sinx}=\mathrm{}\frac{3}{5}\\ \mathrm{Now},\mathrm{cos}\mathrm{x}=\sqrt{1-{\mathrm{sin}}^{2}\mathrm{x}}\\ =\mathrm{}\sqrt{1-{\left(\frac{3}{5}\right)}^{2}}\\ =\mathrm{}\sqrt{\frac{16}{25}}\mathrm{}=\mathrm{}\frac{4}{5}\end{array}$

Q.2

${\text{Solvetan}}^{-1}2\mathrm{x}+{\mathrm{tan}}^{-1}3\mathrm{x}=\frac{\mathrm{\pi }}{4}$

Ans

$\begin{array}{l}{\mathrm{tan}}^{-1}2\mathrm{x}+{\mathrm{tan}}^{-1}3\mathrm{x}\mathrm{}=\mathrm{}\frac{\mathrm{\pi }}{4}\\ ⇒\mathrm{}{\mathrm{tan}}^{-1}\left[\frac{2\mathrm{x}+3\mathrm{x}}{1-2\mathrm{x}×3\mathrm{x}}\right]\mathrm{}=\mathrm{}\frac{\mathrm{\pi }}{4}\mathrm{}\left[\therefore {\mathrm{tan}}^{-1}\mathrm{x}\mathrm{}+{\mathrm{tan}}^{-1}\mathrm{y}\mathrm{}=\mathrm{}{\mathrm{tan}}^{-1}\left[\frac{\mathrm{x}+\mathrm{y}}{1-\mathrm{xy}}\right]\right]\\ ⇒\mathrm{}{\mathrm{tan}}^{-1}\left[\frac{5\mathrm{x}}{1-6{\mathrm{x}}^{2}}\right]\mathrm{}=\mathrm{}\frac{\mathrm{\pi }}{4}\\ \therefore \frac{5\mathrm{x}}{1-6{\mathrm{x}}^{2}}\mathrm{}=\mathrm{}\mathrm{tan}\mathrm{}\frac{\mathrm{\pi }}{4}\\ ⇒\frac{5\mathrm{x}}{1-6{\mathrm{x}}^{2}}\mathrm{}=\mathrm{}1\\ \mathrm{or}5\mathrm{x}\mathrm{}=\mathrm{}1\mathrm{}-\mathrm{}6{\mathrm{x}}^{2}\\ ⇒6{\mathrm{x}}^{2}+5\mathrm{x}-1\mathrm{}=\mathrm{}0\\ \mathrm{i}.\mathrm{e}.\left(6\mathrm{x}-1\right)\left(\mathrm{x}+1\right)\mathrm{}=\mathrm{}0\\ \mathrm{Which}\mathrm{gives},\mathrm{x}=\frac{1}{6},\mathrm{}\mathrm{x}\mathrm{}=\mathrm{}-1\\ \mathrm{Since}\mathrm{x}= -1\mathrm{does}\mathrm{not}\mathrm{satisfy}\mathrm{the}\mathrm{equation}\mathrm{as}\mathrm{the}\mathrm{L}.\mathrm{H}.\mathrm{S}.\mathrm{of}\mathrm{the}\mathrm{equation}\\ \mathrm{becomes}\mathrm{negative},\mathrm{x}=\frac{1}{6}\mathrm{is}\mathrm{the}\mathrm{only}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{equation}.\end{array}$

Q.3

${\text{Prove that tan}}^{-1}\left(\frac{\sqrt{1+\mathrm{x}}-\sqrt{1-\mathrm{x}}}{\sqrt{1+\mathrm{x}+\sqrt{1-\mathrm{x}}}}\right)\mathrm{}=\mathrm{}\frac{\mathrm{\pi }}{4}\mathrm{}-\mathrm{}\frac{1}{2}\mathrm{}{\mathrm{cos}}^{-1}\mathrm{x}$

Ans

$\begin{array}{l}\mathrm{Here},\mathrm{L}.\mathrm{H}.\mathrm{S}. ={\mathrm{tam}}^{-1}\left(\frac{\sqrt{1+\mathrm{x}}-\sqrt{1-\mathrm{x}}}{\sqrt{1+\mathrm{x}}+\mathrm{ }\sqrt{1-\mathrm{x}}}\right)\\ \mathrm{Let}\mathrm{x}=\mathrm{cos}2\mathrm{\theta }\mathrm{}⇒\mathrm{}\mathrm{\theta }=\frac{1}{2}\mathrm{}{\mathrm{cos}}^{-1}\mathrm{x}\\ \mathrm{Then}\mathrm{L}.\mathrm{H}.\mathrm{S}.={\mathrm{tan}}^{-1}\mathrm{}\left[\frac{\sqrt{1+\mathrm{cos}2\mathrm{\theta }}-\sqrt{1-\mathrm{cos}2\mathrm{\theta }}}{\sqrt{1+\mathrm{cos}2\mathrm{\theta }}+\sqrt{1-\mathrm{cos}2\mathrm{\theta }}}\right]\\ =\mathrm{ }{\mathrm{tan}}^{-1}\left[\frac{\sqrt{2{\mathrm{cos}}^{2}\mathrm{\theta }}-\sqrt{2{\mathrm{sin}}^{2}\mathrm{\theta }}}{\sqrt{2{\mathrm{cos}}^{2}\mathrm{\theta }}+\sqrt{2{\mathrm{sin}}^{2}\mathrm{\theta }}}\right]\\ =\mathrm{ }{\mathrm{tan}}^{-1}\left[\frac{\sqrt{2}\mathrm{cos\theta }-\sqrt{2}\mathrm{sin\theta }}{\sqrt{2}\mathrm{cos\theta }+\sqrt{2}\mathrm{sin\theta }}\right]\\ =\mathrm{ }{\mathrm{tan}}^{-1}\left[\frac{\mathrm{cos\theta }-\mathrm{sin\theta }}{\mathrm{cos\theta }+\mathrm{sin\theta }}\right]\\ =\mathrm{ }{\mathrm{tan}}^{-1}\left[\frac{1-\mathrm{tan\theta }}{1+\mathrm{tan\theta }}\right]\\ =\mathrm{ }{\mathrm{tan}}^{-1}\left[\frac{\mathrm{tan}\frac{\mathrm{\pi }}{4}-\mathrm{tan\theta }}{1+\mathrm{tan}\frac{\mathrm{\pi }}{4}\mathrm{tan\theta }}\right]\\ =\mathrm{ }{\mathrm{tan}}^{-1}\left[{\mathrm{tan}}^{-1}\left(\frac{\mathrm{\pi }}{4}-\mathrm{\theta }\right)\right]\\ =\mathrm{}\frac{\mathrm{\pi }}{4}\mathrm{}-\mathrm{\theta }\\ =\mathrm{}\frac{\mathrm{\pi }}{4}-\frac{1}{2}{\mathrm{cos}}^{-1}\mathrm{x}\mathrm{}=\mathrm{R}.\mathrm{H}.\mathrm{S}\mathrm{Proved}\end{array}$

Q.4

${\text{Prove that cot}}^{-1}\left[\frac{\sqrt{1+\mathrm{sinx}}+\sqrt{1-\mathrm{sinx}}}{\sqrt{1+\mathrm{sinx}}-\sqrt{1-\mathrm{sinx}}}\right]\mathrm{}=\mathrm{}\frac{\mathrm{x}}{2},\mathrm{}×\mathrm{}\in \mathrm{}\left(0,\mathrm{}\frac{\mathrm{\pi }}{4}\right).$

Ans

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.\\ ={\mathrm{cot}}^{-1}\left[\frac{\sqrt{1+\mathrm{sinx}}+\sqrt{1-\mathrm{sinx}}}{\sqrt{1+\mathrm{sinx}}-\sqrt{1-\mathrm{sinx}}}\right]\\ =\mathrm{}{\mathrm{cot}}^{-1}\left[\frac{\sqrt{1+\mathrm{sinx}}+\sqrt{1-\mathrm{sinx}}}{\sqrt{1+\mathrm{sinx}}-\sqrt{1-\mathrm{sinx}}}\mathrm{}×\frac{\sqrt{1+\mathrm{sinx}}+\sqrt{1+\mathrm{sinx}}}{\sqrt{1+\mathrm{sinx}}+\sqrt{1-\mathrm{sinx}}}\mathrm{}\right]\\ =\mathrm{}{\mathrm{cot}}^{-1\mathrm{}}\left[\frac{{\left(\sqrt{1+\mathrm{sinx}}+\sqrt{1-\mathrm{sinx}}\right)}^{2}}{1+\mathrm{sinx}-1+\mathrm{sinx}}\right]\\ =\mathrm{}{\mathrm{cot}}^{-1\mathrm{}}\left[\frac{1+\overline{)\mathrm{sinx}}+1-\overline{)\mathrm{sinx}}+2\sqrt{1-{\mathrm{sin}}^{2}\mathrm{x}}}{2\mathrm{sinx}}\right]\\ =\mathrm{}{\mathrm{cot}}^{-1}\mathrm{}\left[\frac{\overline{)2}\left(1+\mathrm{cos}\mathrm{}\mathrm{x}\right)}{\overline{)2}\mathrm{}\mathrm{sin}\mathrm{}\mathrm{x}}\right]\\ =\mathrm{}{\mathrm{cot}}^{-1}\mathrm{}\left[\frac{\overline{)2}{\mathrm{cos}}^{2}\frac{\mathrm{x}}{2}}{\overline{)2}\mathrm{}\mathrm{sin}\mathrm{}\frac{\mathrm{x}}{2}\mathrm{cos}\mathrm{ }\frac{\mathrm{x}}{2}}\right]\\ =\mathrm{}{\mathrm{cot}}^{-1}\mathrm{}\left[\mathrm{cot}\frac{\mathrm{x}}{2}\right]\\ =\mathrm{}\frac{\mathrm{x}}{2}\mathrm{}=\mathrm{R}.\mathrm{H}.\mathrm{S}\end{array}$

Q.5

${\text{Show that sin}}^{-1}\frac{12}{13}+{\mathrm{cos}}^{-1}\frac{4}{5}+{\mathrm{tan}}^{-1}\frac{63}{16}=\mathrm{}\mathrm{\pi }$

Ans

$\begin{array}{l}{\mathrm{sin}}^{-1}\frac{12}{13}\mathrm{}=\mathrm{x},{\mathrm{cos}}^{-1}\frac{4}{5}\mathrm{}=\mathrm{}\mathrm{y}{\mathrm{tan}}^{-1}\frac{63}{16}=\mathrm{}\mathrm{z}\\ ⇒{\mathrm{sin}}^{}\mathrm{x}\mathrm{}=\frac{12}{13},\mathrm{cos}\mathrm{}\mathrm{y}\mathrm{}=\frac{4}{5}\mathrm{}\mathrm{and}\mathrm{tan}\mathrm{}\mathrm{z}\mathrm{}\frac{63}{16}\\ \therefore \mathrm{cos}\mathrm{}\mathrm{x}\mathrm{}=\frac{5}{13},\mathrm{sin}\mathrm{}\mathrm{y}\mathrm{}=\frac{3}{4},\mathrm{}\mathrm{tanx}\mathrm{}=\mathrm{}\frac{12}{5}\mathrm{}\mathrm{and}\mathrm{tan}\mathrm{}\mathrm{y}\mathrm{}\frac{3}{4}\\ \mathrm{Now},\mathrm{tan}\left(\mathrm{x}+\mathrm{y}\right)\mathrm{}=\frac{\mathrm{tanx}+\mathrm{tany}}{1-\mathrm{tan}\mathrm{}×\mathrm{}\mathrm{tany}}=\mathrm{}\frac{\frac{12}{5}+\frac{3}{4}}{1-\frac{12}{5}×\frac{3}{4}}\mathrm{}=\mathrm{}\frac{-63}{16}\\ ⇒\mathrm{}\mathrm{tan}\left(\mathrm{x}+\mathrm{y}\right)\mathrm{}=-\mathrm{tanz}\\ ⇒\mathrm{}\mathrm{tan}\left(\mathrm{x}+\mathrm{y}\right)\mathrm{}=\mathrm{tan}\left(\mathrm{\pi }-\mathrm{z}\right)\\ \therefore \mathrm{x}+\mathrm{y}\mathrm{}=\mathrm{}\mathrm{\pi }-\mathrm{z}\\ ⇒\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{}\mathrm{\pi }\end{array}$

Q.6

$\begin{array}{l}\mathrm{Solve}\mathrm{the}\mathrm{following}\mathrm{equation}:\\ {\mathrm{tan}}^{-1}\left(\frac{\mathrm{x}-1}{\mathrm{x}-2}\right)\mathrm{}+{\mathrm{tan}}^{-1}\left(\frac{\mathrm{x}+1}{\mathrm{x}+2}\right)\mathrm{}=\mathrm{}\frac{\mathrm{\pi }}{4}\end{array}$

Ans

$\begin{array}{l}{\mathrm{tan}}^{-1}\left(\frac{\mathrm{x}-1}{\mathrm{x}-2}\right)+{\mathrm{tan}}^{-1}\left(\frac{\mathrm{x}+1}{\mathrm{x}+2}\right)\mathrm{}=\mathrm{}\frac{\mathrm{\pi }}{4}\\ ⇒\mathrm{}{\mathrm{tan}}^{-1}\mathrm{}\left(\frac{\frac{\mathrm{x}-1}{\mathrm{x}-2}+\frac{\mathrm{x}+1}{\mathrm{x}+2}}{1-\frac{\mathrm{x}-1}{\mathrm{x}-2}×\frac{\mathrm{x}+1}{\mathrm{x}+2}}\right)\mathrm{}=\mathrm{}\frac{\mathrm{\pi }}{4}\\ ⇒\mathrm{}\frac{{\mathrm{x}}^{2}+\overline{)\mathrm{x}}-2+{\mathrm{x}}^{2}-\overline{)\mathrm{x}}-2}{{\overline{)\mathrm{x}}}^{2}-4-{\overline{)\mathrm{x}}}^{2}+1}=\mathrm{}1\\ ⇒\frac{2{\mathrm{x}}^{2}-4}{-3}=\mathrm{}1\\ ⇒2{\mathrm{x}}^{2}-4=\mathrm{}-3\\ ⇒{\mathrm{x}}^{2}=\mathrm{}\frac{1}{2}\\ ⇒\mathrm{x}=\mathrm{}±\frac{1}{\sqrt{2}}\end{array}$

Q.7

$\mathrm{Evaluate}\mathrm{tan}\frac{1}{2}\mathrm{}\left[{\mathrm{sin}}^{-1}\frac{2\mathrm{x}}{1+{\mathrm{x}}^{2}}+\mathrm{}{\mathrm{cos}}^{-1}\frac{1-{\mathrm{y}}^{2}}{1+{\mathrm{y}}^{2}}\right],\left|\mathrm{x}\right|\mathrm{}<\mathrm{}1,\mathrm{}\mathrm{y}\mathrm{}>\mathrm{}0\mathrm{and}\mathrm{xy}> 1$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{x}=\mathrm{tan}\mathrm{\theta }\\ ⇒\mathrm{}\mathrm{\theta }{\mathrm{tan}}^{-1}\mathrm{x}\mathrm{}\mathrm{and}=\mathrm{}{\mathrm{tan}}^{-1}\mathrm{y}\mathrm{then},\\ \mathrm{tan}\frac{1}{2}\left[{\mathrm{sin}}^{-1}\frac{2\mathrm{x}}{1+{\mathrm{x}}^{2}}+{\mathrm{cos}}^{-1}\frac{1-{\mathrm{y}}^{2}}{1+{\mathrm{y}}^{2}}\right]\\ =\mathrm{tan}\frac{1}{2}\left[{\mathrm{sin}}^{-1}\left(\frac{2\mathrm{tan\theta }}{1+{\mathrm{tan}}^{2}\mathrm{\theta }}\right)+{\mathrm{cos}}^{-1}\left(\frac{1-{\mathrm{tan}}^{2}}{1+{\mathrm{tan}}^{2}}\right)\right]\\ =\mathrm{tan}\frac{1}{2}\left[{\mathrm{sin}}^{-1}\left(\mathrm{sin}2\mathrm{\theta }\right)+{\mathrm{cos}}^{-1}\left(\mathrm{cos}2\right)\right]\\ =\mathrm{tan}\mathrm{}\frac{1}{2}\left(2\mathrm{\theta }+2\right)\\ =\mathrm{tan}\left(\mathrm{\theta }+\right)\end{array}$

Q.8

$\mathrm{Show}\mathrm{that}{\mathrm{sin}}^{-1}\frac{3}{5}-{\mathrm{sin}}^{-1}\frac{8}{17}\mathrm{}=\mathrm{}{\mathrm{cos}}^{-1}\frac{84}{85}$

Ans

$\begin{array}{l}\mathrm{Let}{\mathrm{sin}}^{-1}\frac{3}{5}=\mathrm{x}\mathrm{and}{\mathrm{sin}}^{-1}\frac{8}{17}\mathrm{}=\mathrm{}\mathrm{y}\mathrm{}\\ ⇒{\mathrm{sin}}^{-1}\frac{3}{5}=\mathrm{sinx}\mathrm{and}\frac{8}{17}\mathrm{}=\mathrm{}\mathrm{siny}\\ \mathrm{as}\mathrm{cosx}=\sqrt{1-{\mathrm{sin}}^{-2}\mathrm{x}}\\ =\mathrm{}\sqrt{1-\frac{9}{25}}\mathrm{}=\mathrm{}\frac{4}{5}\\ \mathrm{and}\mathrm{cos}\mathrm{y}=\sqrt{1-{\mathrm{sin}}^{2}\mathrm{y}}\\ =\mathrm{}\sqrt{1-\frac{64}{289}}\mathrm{}=\mathrm{}\frac{15}{17}\\ \mathrm{Now},\mathrm{cos}\left(\mathrm{x}-\mathrm{y}\right)\mathrm{}=\mathrm{}\mathrm{cosx}\mathrm{}\mathrm{cosy}\mathrm{}+\mathrm{}\mathrm{sin}\mathrm{}\mathrm{x}\mathrm{}\mathrm{siny}\\ =\mathrm{}\frac{4}{5}×\frac{15}{17}+\frac{3}{5}×\frac{8}{17}\\ \mathrm{cos}\left(\mathrm{x}-\mathrm{y}\right)\mathrm{}=\mathrm{}\frac{84}{85}\\ ⇒\mathrm{x}-\mathrm{y}\mathrm{}=\mathrm{}{\mathrm{cos}}^{-1}\frac{84}{85}\end{array}$

Q.9

$\mathrm{Evaluate}{\mathrm{tan}}^{–1}\left(\mathrm{1}\right)+{\mathrm{cos}}^{–1}\left(\frac{–1}{\mathrm{2}}\right)+{\mathrm{sin}}^{–1}\left(\frac{–1}{\mathrm{2}}\right)$

Ans

$\begin{array}{l}{\mathrm{tan}}^{-1}\left(\mathrm{1}\right)+{\mathrm{cos}}^{–1}\left(\frac{–1}{\mathrm{2}}\right)+{\mathrm{sin}}^{–1}\left(\frac{–1}{\mathrm{2}}\right)\\ {\mathrm{tan}}^{-1}\left(\mathrm{tan}\frac{\mathrm{\pi }}{4}\right)+{\mathrm{cos}}^{–1}\left(-\mathrm{cos}\frac{\mathrm{}\mathrm{\pi }}{3}\right)+{\mathrm{sin}}^{–1}\left(-\mathrm{sin}\frac{\mathrm{\pi }}{6}\right)\\ =\mathrm{}\frac{\mathrm{\pi }}{4}+{\mathrm{cos}}^{–1}\left(\mathrm{cos}\left(\mathrm{\pi }-\frac{\mathrm{}\mathrm{\pi }}{3}\right)\right)+{\mathrm{sin}}^{–1}\left[\mathrm{sin}\left(\frac{-\mathrm{\pi }}{6}\right)\right]\\ \mathrm{we}\mathrm{know}\mathrm{range}\mathrm{of}\\ {\mathrm{tan}}^{-1}\mathrm{}\to \mathrm{}\left(\frac{-\mathrm{\pi }}{2},\mathrm{}\frac{\mathrm{}\mathrm{\pi }}{2}\right),\\ {\mathrm{cos}}^{-1}\mathrm{}\to \mathrm{}\left[0,\mathrm{}\mathrm{\pi }\right]\mathrm{}\mathrm{and}\mathrm{}{\mathrm{sin}}^{-1}\mathrm{}\to \mathrm{}\left[\frac{-\mathrm{\pi }}{2},\mathrm{}\frac{\mathrm{}\mathrm{\pi }}{2}\right]\\ \therefore \mathrm{}{\mathrm{tan}}^{-1}\left(1\right)+{\mathrm{cos}}^{-1}\left(\frac{-1}{2}\right)\mathrm{}+\mathrm{}{\mathrm{sin}}^{-1}\left(\frac{-1}{2}\right)\\ =\frac{\mathrm{\pi }}{4}\mathrm{}+\mathrm{}\frac{2\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{6}\\ \mathrm{}=\mathrm{}\frac{3\mathrm{\pi }+8\mathrm{\pi }-2\mathrm{\pi }}{12}\mathrm{}=\mathrm{}\frac{9\mathrm{\pi }}{12}\mathrm{}=\mathrm{}\frac{3\mathrm{\pi }}{4}\end{array}$

Q.10

$\mathrm{Write}\mathrm{the}\mathrm{function}{\mathrm{tan}}^{-1}\left(\frac{\mathrm{cosx}-\mathrm{sinx}}{\mathrm{cosx}+\mathrm{sinx}}\right)\mathrm{}\mathrm{in}\mathrm{the}\mathrm{simplest}\mathrm{form}.$

Ans

$\begin{array}{l}{\mathrm{tan}}^{-1}\left(\frac{\mathrm{cosx}-\mathrm{sinx}}{\mathrm{cosx}+\mathrm{sinx}}\right)\\ ={\mathrm{tan}}^{-1}\left(\frac{\frac{\mathrm{cosx}}{\mathrm{cosx}}-\frac{\mathrm{sin}\mathrm{}\mathrm{x}}{\mathrm{cos}\mathrm{}\mathrm{x}}}{\frac{\mathrm{cosx}}{\mathrm{cosx}}+\frac{\mathrm{sinx}}{\mathrm{cosx}}}\right)\\ ={\mathrm{tan}}^{-1}\left(\frac{1-\mathrm{tanx}}{1+\mathrm{tanx}}\right)\\ ={\mathrm{tan}}^{-1}\left(\frac{\mathrm{tan}\frac{\mathrm{\pi }}{4}-\mathrm{tanx}}{1+\mathrm{tan}\frac{\mathrm{\pi }}{4}\mathrm{tanx}}\right)\\ \mathrm{}={\mathrm{tan}}^{-1}\left(\mathrm{tan}\left(\frac{\mathrm{\pi }}{4}-\mathrm{x}\right)\right)\left[\therefore \mathrm{}\mathrm{tan}\left(\mathrm{A}-\mathrm{B}\right)=\mathrm{}\frac{\mathrm{tanA}–\mathrm{tanB}}{1+\mathrm{tanA}\mathrm{}\mathrm{tanB}}\right]\\ =\mathrm{}\frac{\mathrm{\pi }}{4}-\mathrm{x}\end{array}$

Q.11

${\text{Express tan}}^{\text{-1}}\left(\frac{\text{cos x}}{\text{1-sin x}}\right)\text{,}\frac{\text{π}}{\text{2}}\text{<\hspace{0.17em}x<}\frac{\text{π}}{\text{2}}\text{in the simplest form.}$

Ans

$\begin{array}{l}{\mathrm{tan}}^{–1}\left(\frac{\mathrm{cos}\mathrm{x}}{1–\mathrm{sin}\mathrm{x}}\right)\mathrm{}={\mathrm{tan}}^{–1}\left[\frac{\mathrm{sin}\left(\frac{\mathrm{\pi }}{2}-\mathrm{x}\right)}{1-\mathrm{cos}\mathrm{}\left(\frac{\mathrm{\pi }}{2}-\mathrm{x}\right)}\right]\\ =\mathrm{}{\mathrm{tan}}^{-1}\left[\frac{\mathrm{sin}\left(\frac{\mathrm{\pi }-2\mathrm{x}}{2}\right)}{1-\mathrm{cos}\mathrm{}\left(\frac{\mathrm{\pi }-2\mathrm{x}}{2}\right)}\right]\\ =\mathrm{}{\mathrm{tan}}^{-1}\left[\frac{2\mathrm{sin}\left(\frac{\mathrm{\pi }-2\mathrm{x}}{4}\right)}{2{\mathrm{sin}}^{2}\mathrm{}\left(\frac{\mathrm{\pi }-2\mathrm{x}}{4}\right)}\right]\\ =\mathrm{}{\mathrm{tan}}^{-1}\left[\frac{\mathrm{cos}\left(\frac{\mathrm{\pi }-2\mathrm{x}}{4}\right)}{\mathrm{sin}\mathrm{}\left(\frac{\mathrm{\pi }-2\mathrm{x}}{4}\right)}\right]\\ =\mathrm{}{\mathrm{tan}}^{-1}\left[\mathrm{cot}\left(\frac{\mathrm{\pi }-2\mathrm{x}}{4}\right)\right]\\ =\mathrm{}{\mathrm{tan}}^{-1}\left[\mathrm{tan}\left(\frac{\mathrm{\pi }}{2}-\left[\frac{\mathrm{\pi }-2\mathrm{x}}{4}\right]\right)\right]\\ =\mathrm{}{\mathrm{tan}}^{-1}\left[\mathrm{tan}\left(\frac{\mathrm{\pi }}{4}+\frac{\mathrm{\pi }}{2}\right)\right]\mathrm{}=\mathrm{}\frac{\mathrm{\pi }}{4}+\frac{\mathrm{x}}{2}\end{array}$

Q.12

$\mathrm{Find}\mathrm{the}\mathrm{value}\mathrm{of}{\mathrm{sin}}^{–1}\mathrm{}\left(\mathrm{sin}\frac{2\mathrm{\pi }}{\mathrm{3}}\right)$

Ans

$\begin{array}{l}\mathrm{The}\mathrm{range}\mathrm{of}\mathrm{inverse}\mathrm{function}\mathrm{is}\left[\frac{-\mathrm{\pi }}{2},\mathrm{}\frac{-\mathrm{\pi }}{2}\right].\\ \mathrm{Now},{\mathrm{sin}}^{-1}\left[\mathrm{sin}\left(\frac{2\mathrm{\pi }}{3}\right)\right]\mathrm{}={\mathrm{sin}}^{-1}\left[\mathrm{sin}\left(\mathrm{\pi }-\frac{\mathrm{\pi }}{3}\right)\right]\\ ={\mathrm{sin}}^{-1}\left[\mathrm{sin}\left(\frac{\mathrm{\pi }}{3}\right)\right]\\ =\mathrm{}\frac{\mathrm{\pi }}{3}\end{array}$

Q.13

$\mathrm{Write}{\mathrm{tan}}^{-1}\left(\frac{1}{\sqrt{{\mathrm{x}}^{2}-1}}\right),\mathrm{}\left|\mathrm{x}\right|\mathrm{}>\mathrm{}1\mathrm{in}\mathrm{the}\mathrm{simplest}\mathrm{form}.$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{x}=\mathrm{cosec}\mathrm{\theta }\\ ⇒\mathrm{}\mathrm{\theta }={\mathrm{cosec}}^{-1}\mathrm{x}\\ \mathrm{Then},{\mathrm{tan}}^{-1}\left(\frac{1}{\sqrt{{\mathrm{x}}^{2}-1}}\right)={\mathrm{tan}}^{-1}\left(\frac{1}{\sqrt{{\mathrm{cosec}}^{2}-1}}\right)\\ ={\mathrm{tan}}^{-1}\left(\frac{1}{\mathrm{cot\theta }}\right)\\ ={\mathrm{tan}}^{-1}\left(\mathrm{tan\theta }\right)\\ =\mathrm{}\mathrm{\theta }\\ =\mathrm{}{\mathrm{cosec}}^{-1}\mathrm{x}\end{array}$

Q.14

$\mathrm{Prove}\mathrm{that}{\mathrm{tan}}^{-1}\mathrm{x}+{\mathrm{tan}}^{-1}\frac{2\mathrm{x}}{1-{\mathrm{x}}^{2}}={\mathrm{tan}}^{-1}\left(\frac{3\mathrm{x}-{\mathrm{x}}^{3}}{1-3{\mathrm{x}}^{2}}\right)$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{x}=\mathrm{tan}\mathrm{\theta }\\ \mathrm{Now},\\ \mathrm{L}.\mathrm{H}.\mathrm{S}. ={\mathrm{tan}}^{-1}\mathrm{x}+\mathrm{}{\mathrm{tan}}^{-1}\frac{2\mathrm{x}}{1-{\mathrm{x}}^{2}}\\ =\mathrm{}{\mathrm{tan}}^{-1}\left(\mathrm{tan\theta }\right)+\mathrm{}{\mathrm{tan}}^{-1}\left(\frac{2\mathrm{}\mathrm{tan\theta }}{1-{\mathrm{tan}}^{2}\mathrm{\theta }}\right)\\ =\mathrm{}\mathrm{\theta }\mathrm{}+\mathrm{}{\mathrm{tan}}^{-1}\mathrm{}\left(\mathrm{tan}2\mathrm{\theta }\right)\\ =\mathrm{}\mathrm{\theta }\mathrm{}+\mathrm{}2\mathrm{\theta }\mathrm{}=\mathrm{}3\mathrm{\theta }.\dots ..\left(1\right)\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.={\mathrm{tan}}^{-1}\left(\frac{3\mathrm{}\mathrm{tan\theta }-{\mathrm{tan}}^{3}\mathrm{\theta }}{1-3\mathrm{}{\mathrm{tan}}^{2}\mathrm{\theta }}\right)\\ ={\mathrm{tan}}^{-1}\left(3\mathrm{}\mathrm{tan\theta }\right)\mathrm{}=\mathrm{}3\mathrm{}\mathrm{\theta }.\dots .\left(2\right)\\ \mathrm{by}\left(1\right)\mathrm{}\mathrm{and}\mathrm{}\left(2\right)\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{proved}\end{array}$

Q.15

$\mathrm{Show}\mathrm{that}:{\mathrm{tan}}^{-1}\frac{1}{2}+{\mathrm{tan}}^{-1}\frac{2}{11}={\mathrm{tan}}^{-1}\frac{3}{4}$

Ans

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.={\mathrm{tan}}^{-1}\left[\frac{\frac{1}{2}+\frac{2}{11}}{1-\frac{1}{2}×\frac{2}{11}}\right]\left[{\mathrm{tan}}^{-1}\mathrm{x}+\mathrm{}{\mathrm{tan}}^{-1}\mathrm{y}\mathrm{}=\mathrm{}{\mathrm{tan}}^{-1}\left(\frac{\mathrm{x}+\mathrm{y}}{1-\mathrm{xy}}\right)\right]\\ =\mathrm{}{\mathrm{tan}}^{-1}\frac{15}{20}\\ =\mathrm{}{\mathrm{tan}}^{-1}\frac{3}{4}\mathrm{}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.16

$\begin{array}{l}\text{Find the principal value of}\\ \left(\text{i}\right){\text{sin}}^{\text{-1}}\left(\frac{\text{-1}}{\text{2}}\right)\\ \left(\text{ii}\right){\text{cot}}^{\text{-1}}\left(\frac{\text{-1}}{\sqrt{\text{3}}}\right)\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right){\mathrm{sin}}^{–1}\left(\frac{–1}{\mathrm{2}}\right)\mathrm{}=\mathrm{}\mathrm{y}\\ ⇒\mathrm{sin}\mathrm{}\mathrm{y}\mathrm{}=\mathrm{}\frac{-1}{2}\\ ⇒\mathrm{sin}\mathrm{}\mathrm{y}\mathrm{}=\mathrm{}-\mathrm{sin}\mathrm{}\frac{\mathrm{\pi }}{6}\\ ⇒\mathrm{sin}\mathrm{}\mathrm{y}\mathrm{}=\mathrm{}-\mathrm{sin}\mathrm{}\left(\frac{-\mathrm{\pi }}{6}\right)\mathrm{}\mathrm{or}\mathrm{}\mathrm{y}\mathrm{}=\mathrm{}\frac{-\mathrm{\pi }}{6}\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{inverse}\mathrm{sine}\mathrm{function}\mathrm{is}\left[\frac{-\mathrm{\pi }}{2},\mathrm{}\frac{\mathrm{\pi }}{2}\right]\\ \therefore {\mathrm{sin}}^{-1}\mathrm{}\left(\frac{-\mathrm{\pi }}{6}\right)\mathrm{}=\mathrm{}\frac{-\mathrm{\pi }}{6}\\ \left(\mathrm{ii}\right)\mathrm{Let}{\mathrm{cot}}^{-1}\mathrm{}\left(\frac{-1}{\sqrt{3}}\right)\mathrm{}=\mathrm{}\mathrm{y}\\ ⇒\mathrm{coty}\mathrm{}=\mathrm{}\frac{-1}{\sqrt{3}}\\ ⇒\mathrm{coty}\mathrm{}=\mathrm{}\frac{-1}{\sqrt{3}}\\ ⇒\mathrm{coty}\mathrm{}=\mathrm{}\mathrm{cot}\mathrm{}\left(\mathrm{\pi }-\frac{\mathrm{\pi }}{3}\right)\\ ⇒\mathrm{y}\mathrm{}=\mathrm{}\frac{2\mathrm{\pi }}{3}\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{range}\mathrm{of}\mathrm{inverse}\mathrm{contangent}\mathrm{function}\left(0,\mathrm{}\mathrm{\pi }\right)\\ \therefore \mathrm{}{\mathrm{cot}}^{-1}\mathrm{}\left(\frac{-1}{\sqrt{3}}\right)\mathrm{}=\mathrm{}\frac{2\mathrm{\pi }}{3}\end{array}$

Q.17

$\begin{array}{l}\mathrm{Is}\mathrm{the}\mathrm{following}\mathrm{statement}\mathrm{true}?\\ {\mathrm{cos}}^{–1}\mathrm{x}={\left(\mathrm{cosx}\right)}^{–1}\end{array}$

Ans

$\mathrm{No},{\mathrm{cos}}^{–1}\mathrm{x}¹\ne \mathrm{}{\left(\mathrm{cos}\mathrm{x}\right)}^{–1}=\frac{\mathrm{1}}{\mathrm{cos}\mathrm{x}}$

Q.18

$\mathrm{Find}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{of}{\mathrm{cos}}^{–1}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)$

Ans

$\begin{array}{l}\mathrm{Let}{\mathrm{cos}}^{–1}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\mathrm{}=\mathrm{y}\\ ⇒\mathrm{}\mathrm{cos}\mathrm{}\mathrm{y}\mathrm{}=\mathrm{}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\\ \mathrm{Range}\mathrm{of}{\mathrm{cos}}^{-1}\mathrm{}\mathrm{is}\mathrm{}\left[0,\mathrm{}\mathrm{\pi }\right]\\ \mathrm{cos}\frac{\mathrm{\pi }}{6}\mathrm{}=\mathrm{}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\\ ⇒\mathrm{}{\mathrm{cos}}^{-1}\mathrm{}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{}=\mathrm{}\frac{\mathrm{\pi }}{6}\end{array}$

Q.19

$\mathrm{If}{\mathrm{sin}}^{-1}\mathrm{}\mathrm{x}\mathrm{}=\mathrm{}\mathrm{y},\mathrm{than}\mathrm{what}\mathrm{will}\mathrm{be}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{y}?$

Ans

$\begin{array}{l}\mathrm{principal}\mathrm{value}\mathrm{of}{\mathrm{sin}}^{-1}\mathrm{lies}\mathrm{between}\left[-\frac{\mathrm{\pi }}{2},\mathrm{}\frac{\mathrm{\pi }}{2}\right]\\ -\frac{\mathrm{\pi }}{2}\mathrm{}\le \frac{\mathrm{\pi }}{2}\mathrm{}\mathrm{y}\mathrm{}\le \mathrm{}\frac{\mathrm{\pi }}{2}\end{array}$

Q.20

${\text{Show that sin}}^{-1}\left(2\mathrm{x}\sqrt{1-{\mathrm{x}}^{2}}\right)\mathrm{}=\mathrm{}2{\mathrm{sin}}^{-1}\mathrm{x}$

Ans

$\begin{array}{l}{\mathrm{sin}}^{-1}\left(2\mathrm{x}\sqrt{1-{\mathrm{x}}^{2}}\right)=\mathrm{}2{\mathrm{sin}}^{-1}\mathrm{x}\\ \mathrm{let}\mathrm{x}=\mathrm{sin\theta }\\ {\mathrm{sin}}^{-1}\left(2\mathrm{x}\sqrt{1-{\mathrm{x}}^{2}}\right)=\mathrm{}{\mathrm{sin}}^{-1}\left(2\mathrm{}\mathrm{sin\theta }\mathrm{}\sqrt{1-{\mathrm{sin}}^{2}\mathrm{\theta }}\right)\\ =\mathrm{}{\mathrm{sin}}^{-1}\left(2\mathrm{}\mathrm{sin\theta }\mathrm{}\mathrm{cos\theta }\right)\\ =\mathrm{}{\mathrm{sin}}^{-1}\left(\mathrm{}\mathrm{sin}2\mathrm{\theta }\right)\\ =\mathrm{}2\mathrm{\theta }\\ =\mathrm{}2{\mathrm{sin}}^{-1}\mathrm{x}\end{array}$

Q.21

$\mathrm{Find}\mathrm{the}\mathrm{value}\mathrm{of}{\mathrm{cos}}^{-1}\left(\mathrm{cos}\frac{13\mathrm{\pi }}{6}\right)\mathrm{.}$

Ans

$\begin{array}{l}{\mathrm{cos}}^{-1}\left(\mathrm{cos}\frac{13\mathrm{\pi }}{6}\right)\mathrm{}=\mathrm{}\frac{13\mathrm{\pi }}{6}\\ \mathrm{But}\frac{13\mathrm{\pi }}{6}\mathrm{}\notin \mathrm{}\left[0,\mathrm{}\mathrm{\pi }\right]\\ \mathrm{cos}\left(\frac{13\mathrm{\pi }}{6}\right)\mathrm{}=\mathrm{cos}\left(2\mathrm{\pi }+\frac{\mathrm{\pi }}{6}\right)\mathrm{}=\mathrm{}\mathrm{cos}\mathrm{}\left(\frac{\mathrm{\pi }}{6}\right)\\ \therefore {\mathrm{cos}}^{-1}\mathrm{}\left(\mathrm{cos}\frac{13\mathrm{\pi }}{6}\right)\mathrm{}=\mathrm{}\frac{\mathrm{\pi }}{6}\end{array}$

Q.22

$\mathrm{Find}\mathrm{the}\mathrm{value}\mathrm{of}{\mathrm{tan}}^{–1}\left(\frac{\mathrm{2}}{\mathrm{11}}\right)+{\mathrm{tan}}^{–1}\left(\frac{\mathrm{7}}{\mathrm{24}}\right)\mathrm{.}$

Ans

$\begin{array}{l}{\mathrm{tan}}^{–1}\left(\frac{\mathrm{2}}{\mathrm{11}}\right)+{\mathrm{tan}}^{–1}\left(\frac{\mathrm{7}}{\mathrm{24}}\right)\mathrm{.}\\ ={\mathrm{tan}}^{–1}\left(\frac{\frac{\mathrm{2}}{\mathrm{11}}+\frac{7}{24}}{1-\frac{2}{11}×\frac{7}{24}}\right)\left[{\mathrm{tan}}^{-1}\mathrm{x}+{\mathrm{tan}}^{-1}\mathrm{y}\mathrm{}=\mathrm{}{\mathrm{tan}}^{-1}\left(\frac{\mathrm{x}+\mathrm{y}}{1-\mathrm{xy}}\right)\right]\\ =\mathrm{}{\mathrm{tan}}^{-1}\mathrm{}\left(\frac{125}{250}\right)\\ =\mathrm{}{\mathrm{tan}}^{-1}\mathrm{}\left(\frac{1}{2}\right)\end{array}$

Q.23

$\mathrm{Find}\mathrm{the}\mathrm{value}\mathrm{of}{\mathrm{tan}}^{–1}\left(\frac{3{\mathrm{a}}^{\mathrm{2}}\mathrm{x}-{\mathrm{x}}^{\mathrm{3}}}{{\mathrm{a}}^{\mathrm{3}}-3{\mathrm{ax}}^{\mathrm{2}}}\right)\mathrm{.}$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{x}=\mathrm{a}\mathrm{tan}\mathrm{\theta }\\ \mathrm{Then}{\mathrm{tan}}^{–1}\mathrm{}\left(\frac{3{\mathrm{a}}^{\mathrm{2}}\mathrm{x}-{\mathrm{x}}^{\mathrm{3}}}{{\mathrm{a}}^{\mathrm{3}}-3{\mathrm{ax}}^{\mathrm{2}}}\right)={\mathrm{tan}}^{–1}\mathrm{}\left(\frac{3{\mathrm{a}}^{\mathrm{2}}\mathrm{tan\theta }-{\mathrm{a}}^{\mathrm{3}}{\mathrm{tan}}^{\mathrm{3}}\mathrm{\theta }}{{\mathrm{a}}^{\mathrm{3}}-3{\mathrm{a}}^{\mathrm{2}}{\mathrm{tan}}^{\mathrm{2}}\mathrm{\theta }}\right)\\ ={\mathrm{tan}}^{–1}\mathrm{}\left(\frac{3\mathrm{tan\theta }-{\mathrm{tan}}^{\mathrm{3}}\mathrm{\theta }}{\mathrm{1}-3{\mathrm{tan}}^{\mathrm{2}}\mathrm{\theta }}\right)\\ ={\mathrm{tan}}^{–1}\mathrm{}\left(\mathrm{tan}3\mathrm{\theta }\right)\\ =3\mathrm{\theta }=3{\mathrm{tan}}^{–1}\mathrm{}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)\end{array}$

Q.24

$\mathrm{Write}2{\mathrm{tan}}^{–1}\mathrm{x}\mathrm{in}\mathrm{terms}\mathrm{of}{\mathrm{sin}}^{–1}\mathrm{x},{\mathrm{cos}}^{–1}\mathrm{x}\mathrm{and}{\mathrm{tan}}^{–1}\mathrm{x}.$

Ans

$2{\mathrm{tan}}^{–1}\mathrm{x}={\mathrm{sin}}^{–1}\mathrm{}\left(\frac{2\mathrm{x}}{1+{\mathrm{x}}^{2}}\right)={\mathrm{cos}}^{–1}\mathrm{}\left(\frac{1-{\mathrm{x}}^{2}}{1+{\mathrm{x}}^{2}}\right)={\mathrm{tan}}^{–1}\mathrm{}\left(\frac{2\mathrm{x}}{1-{\mathrm{x}}^{2}}\right)$

Q.25

Write the domain and range of tan-1x.

Ans

$\begin{array}{l}\mathrm{Domain}\mathrm{of}{\mathrm{tan}}^{–1}\mathrm{x}=\mathrm{R}\\ \mathrm{Range}\mathrm{of}{\mathrm{ten}}^{–1}\mathrm{x}=\left(-\frac{\mathrm{\pi }}{\mathrm{2}},\frac{\mathrm{\pi }}{\mathrm{2}}\right)\end{array}$

Q.26

Fill in the following blanks:

${\text{a)cos}}^{-\mathrm{1}}\left(-\mathrm{x}\right)=.\dots \mathrm{b}\right){\mathrm{cos}}^{-\mathrm{1}} \frac{\mathrm{1}}{\mathrm{x}}=.\dots .$

Ans

$\begin{array}{l}\mathrm{a}\right){\mathrm{cos}}^{-\mathrm{1}}\left(-\mathrm{x}\right)=\mathrm{\pi }\mathrm{}-{\mathrm{cos}}^{-1}\mathrm{x}\\ \mathrm{b}\right){\mathrm{cos}}^{-\mathrm{1}} \frac{\mathrm{1}}{\mathrm{x}}={\mathrm{sec}}^{-1}\mathrm{x}\end{array}$

Q.27

$\text{Find the value of sin}\left[\frac{\mathrm{\pi }}{3}-{\mathrm{sin}}^{-1}\left(-\frac{1}{2}\right)\right].$

Ans

$\begin{array}{l}\mathrm{sin}\left[\frac{\mathrm{\pi }}{3}-{\mathrm{sin}}^{-1}\left(-\frac{1}{2}\right)\right]\\ =\mathrm{sin}\left[\frac{\mathrm{\pi }}{3}\mathrm{}+\mathrm{}{\mathrm{sin}}^{-1}\left(-\frac{1}{2}\right)\right]\\ =\mathrm{sin}\left[\frac{\mathrm{\pi }}{3}+\frac{\mathrm{\pi }}{6}\right]\\ =\mathrm{sin}\mathrm{}\frac{\mathrm{\pi }}{2}\mathrm{}\\ =\mathrm{}1.\end{array}$

Q.28

$\mathrm{Evaluate}{\mathrm{tan}}^{–1}\mathrm{}\left(\mathrm{tan}\frac{4\mathrm{\pi }}{\mathrm{3}}\right)\mathrm{.}$

Ans

$\begin{array}{l}\mathrm{The}\mathrm{rage}\mathrm{of}\mathrm{inverse}\mathrm{tangent}\mathrm{function}\mathrm{is}\left(\frac{-\mathrm{\pi }}{2},\mathrm{}\frac{\mathrm{\pi }}{2}\mathrm{}\right).\\ \mathrm{Now},{\mathrm{tan}}^{-1}\left[\mathrm{tan}\left(\frac{4\mathrm{\pi }}{3}\right)\right]\mathrm{}=\mathrm{}{\mathrm{tan}}^{-1}\left[\mathrm{tan}\left(\mathrm{\pi }+\frac{\mathrm{\pi }}{3}\right)\right]\mathrm{}\\ =\mathrm{}{\mathrm{tan}}^{-1}\left(\frac{\mathrm{\pi }}{3}\right)\\ =\mathrm{}\frac{\mathrm{\pi }}{3}\end{array}$

Q.29

$\text{Solve sin}\left\{2{\mathrm{cos}}^{-1}\left(\mathrm{cot}\left(2\mathrm{}{\mathrm{tan}}^{-1}\mathrm{x}\right)\right)\right\}\mathrm{}=\mathrm{}0.$

Ans

$\begin{array}{l}\mathrm{Given}\mathrm{equation}\mathrm{is}\\ \mathrm{sin}\left\{2{\mathrm{cos}}^{-1}\left(\mathrm{cot}\left(2\mathrm{}{\mathrm{tan}}^{-1}\mathrm{x}\right)\right)\right\}\mathrm{}=\mathrm{}0\\ ⇒\mathrm{}2{\mathrm{cos}}^{-1}\left(\mathrm{cot}\left(2\mathrm{}{\mathrm{tan}}^{-1}\mathrm{x}\right)\right)\mathrm{}=\mathrm{}{\mathrm{n}}^{\mathrm{\pi }},\mathrm{where}\mathrm{n}\mathrm{is}\mathrm{any}\mathrm{integer}\\ ⇒\mathrm{}{\mathrm{cos}}^{-1}\left(\mathrm{cot}\left(2\mathrm{}{\mathrm{tan}}^{-1}\mathrm{x}\right)\right)\mathrm{}=\mathrm{}{\mathrm{n}}^{\mathrm{\pi }/2}\\ ⇒\mathrm{}{\mathrm{cos}}^{-1}\left(\mathrm{cot}\left(2\mathrm{}{\mathrm{tan}}^{-1}\mathrm{x}\right)\right)\mathrm{}=0,\frac{\mathrm{\pi }}{2},\mathrm{\pi }\\ \left(\mathrm{since}{\mathrm{cos}}^{-1}\mathrm{x}\mathrm{}\mathrm{lies}\mathrm{}\mathrm{in}\mathrm{}\left[0,\mathrm{}\mathrm{\pi }\right]\right)\\ ⇒\mathrm{}\left(\mathrm{cot}\left(2{\mathrm{tan}}^{-1}\mathrm{x}\right)\right)\mathrm{}=\mathrm{}\mathrm{cos}\mathrm{}0,\mathrm{}\mathrm{cos}\mathrm{}\frac{\mathrm{\pi }}{2},\mathrm{}\mathrm{cos}\mathrm{}\mathrm{\pi }\mathrm{}=\mathrm{}1,\mathrm{}0,\mathrm{}-1\\ ⇒\mathrm{}\mathrm{cot}\mathrm{}\left[{\mathrm{tan}}^{-1}\left\{\left(2\mathrm{x}\right)/\left(1-{\mathrm{x}}^{2}\right)\right\}\right]\mathrm{}=\mathrm{}1,\mathrm{}0,\mathrm{}-1\\ ⇒\mathrm{}1/\left[\mathrm{tan}\left\{{\mathrm{tan}}^{-1}\left[\left(2\mathrm{x}\right)/\left(1-{\mathrm{x}}^{2}\right)\right]\right\}\right]=\mathrm{}1,\mathrm{}0,\mathrm{}-1\\ ⇒\mathrm{}\left(1-{\mathrm{x}}^{2}\right)/\left(2\mathrm{x}\right)\mathrm{}=\mathrm{}1,\mathrm{}0,\mathrm{}-1\\ ⇒\mathrm{}\left(1-{\mathrm{x}}^{2}\right)/\left(2\mathrm{x}\right)\mathrm{}=\mathrm{}1,\mathrm{}\left(1-{\mathrm{x}}^{2}\right)/\left(2\mathrm{x}\right)\mathrm{}=0\mathrm{and}\left(1-{\mathrm{x}}^{2}\right)/\left(2\mathrm{x}\right)\mathrm{}=\mathrm{}-1\\ ⇒\mathrm{}\mathrm{x}\mathrm{}=\mathrm{}-1±\mathrm{}\sqrt{2},\mathrm{}\mathrm{x}=\mathrm{}±\mathrm{}1,\mathrm{}\mathrm{x}\mathrm{}=\mathrm{}1\mathrm{}±\mathrm{}\sqrt{2}.\end{array}$

Q.30

${\text{Prove that sin}}^{-1}\frac{12}{13}+{\mathrm{cos}}^{-1}\frac{4}{5}\mathrm{}=\mathrm{}\mathrm{sinh}-1\frac{63}{65}.$

Ans

$\begin{array}{l}{\mathrm{sin}}^{-1}\frac{12}{13}\mathrm{}=\mathrm{}\mathrm{x}\mathrm{and}{\mathrm{cos}}^{-1}\frac{4}{5}\mathrm{}=\mathrm{}\mathrm{y}\\ \therefore \mathrm{}\mathrm{cos}\mathrm{}\mathrm{x}\mathrm{}\frac{5}{13}\mathrm{and}\mathrm{}\mathrm{siny}\mathrm{}=\mathrm{}\frac{3}{5}\\ \mathrm{Now},\mathrm{sin}\left(\mathrm{x}+\mathrm{y}\right)\mathrm{}=\mathrm{}\mathrm{sinx}\mathrm{}\mathrm{cosy}\mathrm{}+\mathrm{}\mathrm{cosx}\mathrm{}\mathrm{siny}\\ =\mathrm{}\frac{12}{13},\mathrm{}\frac{4}{5}+\mathrm{}\frac{5}{13},\mathrm{}\frac{3}{5}\mathrm{}=\mathrm{}\frac{63}{65}\\ ⇒\mathrm{}\mathrm{x}+\mathrm{y}\mathrm{}=\mathrm{}{\mathrm{sin}}^{-1}\frac{63}{65}\\ \mathrm{so},{\mathrm{sin}}^{-1}\mathrm{}\frac{12}{13}\mathrm{}+\mathrm{}{\mathrm{cos}}^{-1}\mathrm{}\frac{4}{5}\mathrm{}=\mathrm{}{\mathrm{sin}}^{-1}\mathrm{}\frac{63}{65}\end{array}$

Q.31

$\mathrm{Prove}\mathrm{that}{\mathrm{tan}}^{–1}\frac{\mathrm{1}}{\mathrm{5}}+{\mathrm{tan}}^{–1}\frac{\mathrm{1}}{\mathrm{6}}={\mathrm{tan}}^{–1}\mathrm{}\frac{\mathrm{11}}{\mathrm{29}}$

Ans

$\begin{array}{l}\mathrm{Prove}\mathrm{that}{\mathrm{tan}}^{–1}\frac{\mathrm{1}}{\mathrm{5}}+{\mathrm{tan}}^{–1}\frac{\mathrm{1}}{\mathrm{6}}={\mathrm{tan}}^{–1}\mathrm{}\frac{\mathrm{11}}{\mathrm{29}}\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.={\mathrm{tan}}^{-1}\frac{1}{5}\mathrm{}+\mathrm{}{\mathrm{tan}}^{-1}\frac{1}{6}\\ =\mathrm{}{\mathrm{tan}}^{-1}\mathrm{}\left[\frac{\frac{1}{5}+\frac{1}{6}}{1-\frac{1}{5}×\frac{1}{6}}\right]\\ =\mathrm{}{\mathrm{tan}}^{-1}\mathrm{}\left[\frac{6+5}{30-1}\right]\\ ={\mathrm{tan}}^{-1}\left[\frac{11}{29}\right]\mathrm{}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Proved}.\end{array}$

Q.32

${\text{Find the principal value of sin}}^{-1}\left(\frac{1}{2}\right).$

Ans

$\begin{array}{l}\mathrm{Let}{\mathrm{sin}}^{-1}\mathrm{}\left(\frac{1}{2}\right)\mathrm{}=\mathrm{\theta }\\ ⇒\mathrm{}\mathrm{sin}\mathrm{\theta }=\frac{1}{2}\\ ⇒\mathrm{sin}\mathrm{\theta }=\mathrm{}\mathrm{sin}\mathrm{ }\left(\frac{\mathrm{\pi }}{6}\right)\\ ⇒\mathrm{sin}\mathrm{}\mathrm{\theta }\mathrm{}=\mathrm{sin}\mathrm{ }\left(\frac{\mathrm{\pi }}{6}\right)\mathrm{}\left(âˆµ\mathrm{}\mathrm{The}\mathrm{range}\mathrm{of}\mathrm{sin}{–}^{-1}\mathrm{x}\mathrm{}\mathrm{is}\mathrm{}\left[\frac{\mathrm{\pi }}{2},\mathrm{}\frac{\mathrm{\pi }}{2}\right]\right)\\ ⇒\mathrm{}\mathrm{\theta }\mathrm{}=\frac{\mathrm{\pi }}{6}\\ \therefore \mathrm{The}\mathrm{principal}\mathrm{value}\mathrm{of}{\mathrm{sin}}^{-1}\mathrm{}\left(\frac{1}{2}\right)\mathrm{}\mathrm{is}\mathrm{}\frac{\mathrm{\pi }}{6}.\end{array}$

Q.33

${\text{Final the principal value of sin}}^{-1}\left(\frac{1}{\sqrt{2}}\right).$

Ans

$\begin{array}{l}\mathrm{Let}{\mathrm{sin}}^{-1}\mathrm{}\left(\frac{1}{\sqrt{2}}\right)\mathrm{}=\mathrm{\theta }\\ ⇒\mathrm{}\mathrm{sin}\mathrm{\theta }=\frac{1}{\sqrt{2}}\\ ⇒\mathrm{cos}\mathrm{\theta }=-\mathrm{}\mathrm{cos}\mathrm{ }\left(\frac{\mathrm{\pi }}{4}\right)\\ ⇒\mathrm{sin}\mathrm{}\mathrm{\theta }\mathrm{}=\mathrm{sin}\mathrm{ }\left(\frac{\mathrm{\pi }}{4}\right)\mathrm{}\left(âˆµ\mathrm{}\mathrm{The}\mathrm{range}\mathrm{of}\mathrm{sin}{–}^{-1}\mathrm{x}\mathrm{}\mathrm{is}\mathrm{}\left[\frac{\mathrm{\pi }}{2},\mathrm{}\frac{\mathrm{\pi }}{2}\right]\right)\\ ⇒\mathrm{}\mathrm{\theta }\mathrm{}=\frac{\mathrm{\pi }}{4}\\ \therefore \mathrm{The}\mathrm{principal}\mathrm{value}\mathrm{of}{\mathrm{sin}}^{-1}\mathrm{}\left(\frac{1}{\sqrt{2}}\right)\mathrm{}\mathrm{is}\mathrm{}\frac{\mathrm{\pi }}{4}.\end{array}$

Q.34

$\mathrm{Find}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{of}{\mathrm{cos}}^{–1}\mathrm{}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)\mathrm{.}$

Ans

$\begin{array}{l}\mathrm{Let}{\mathrm{cos}}^{-1}\mathrm{}\left(-\frac{1}{\sqrt{2}}\right)\mathrm{}=\mathrm{}0\\ ⇒\mathrm{}\mathrm{cos}\mathrm{}\mathrm{\theta }\mathrm{}=\mathrm{}-\mathrm{}\frac{1}{\sqrt{2}}\\ ⇒\mathrm{cos}\mathrm{}\mathrm{\theta }\mathrm{}=-\mathrm{}\mathrm{cos}\mathrm{ }\left(\frac{\mathrm{\pi }}{4}\right)\\ ⇒\mathrm{cos}\mathrm{}\mathrm{\theta }\mathrm{}=\mathrm{cos}\mathrm{ }\left(\frac{\mathrm{\pi }}{4}\right)\mathrm{}\left(âˆµ\mathrm{}\mathrm{The}\mathrm{range}\mathrm{of}\mathrm{cos}{–}^{-1}\mathrm{x}\mathrm{}\mathrm{is}\mathrm{}\left[0,\mathrm{}\mathrm{\pi }\right]\right)\\ ⇒\mathrm{cos}\mathrm{}\mathrm{\theta }\mathrm{}=\mathrm{}\mathrm{cos}\mathrm{ }\left(\frac{3\mathrm{\pi }}{4}\right)\\ ⇒\mathrm{}\mathrm{\theta }\mathrm{}=-\mathrm{}\frac{3\mathrm{\pi }}{4}\\ \therefore \mathrm{The}\mathrm{principal}\mathrm{value}\mathrm{of}{\mathrm{cos}}^{-1}\mathrm{}\left(-\frac{1}{\sqrt{2}}\right)\mathrm{}\mathrm{is}\mathrm{}\frac{3\mathrm{\pi }}{4}.\end{array}$

Q.35

$\mathrm{Evaluate}\mathrm{the}\mathrm{cot}\left({\mathrm{tan}}^{–1}\mathrm{a}+{\mathrm{cot}}^{–1}\mathrm{a}\right)\mathrm{.}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},\\ =\mathrm{cot}\left({\mathrm{tan}}^{–1}\mathrm{a}+{\mathrm{cot}}^{–1}\mathrm{a}\right)\\ =\mathrm{cot}\left(\mathrm{}\frac{\mathrm{\pi }}{\mathrm{2}}\right)\mathrm{}\left[âˆµ{\mathrm{tan}}^{–1}\mathrm{a}+{\mathrm{cot}}^{–1}\mathrm{a}=\frac{\mathrm{\pi }}{\mathrm{2}}\right]\\ =0\end{array}$

Q.36

$\text{Evaluate the sin}\left({\text{sin}}^{\text{-1}}{\text{x+cos}}^{\text{-1}}\text{x}\right)\text{.}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},\\ =\mathrm{sin}\left({\mathrm{sin}}^{–1}\mathrm{x}+{\mathrm{cos}}^{–1}\mathrm{x}\right)\\ =\mathrm{sin}\left(\mathrm{}\frac{\mathrm{\pi }}{\mathrm{2}}\right)\mathrm{}\left[âˆµ{\mathrm{sin}}^{–1}\mathrm{x}+{\mathrm{cos}}^{–1}\mathrm{x}=\frac{\mathrm{\pi }}{\mathrm{2}}\right]\\ =1\end{array}$

Q.37

$\mathrm{Put}\mathrm{in}\mathrm{the}\mathrm{simplest}\mathrm{form}:{\mathrm{tan}}^{–1}\left(\frac{1–\mathrm{cos\theta }}{\mathrm{sin\theta }}\right)\mathrm{.}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},{\mathrm{tan}}^{–1}\left(\frac{1–\mathrm{cos}\mathrm{\theta }}{\mathrm{sin}\mathrm{\theta }}\right)\\ ={\mathrm{tan}}^{–1}\mathrm{}\left(\frac{2{\mathrm{sin}}^{\mathrm{2}}\mathrm{}\frac{\mathrm{\theta }}{\mathrm{2}}}{2\mathrm{sin}\frac{\mathrm{\theta }}{\mathrm{2}}\mathrm{cos}\frac{\mathrm{\theta }}{\mathrm{2}}}\right)\\ ={\mathrm{tan}}^{–1}\mathrm{}\left(\frac{\mathrm{sin}\frac{\mathrm{\theta }}{\mathrm{2}}}{\mathrm{cos}\frac{\mathrm{\theta }}{\mathrm{2}}}\right)\\ ={\mathrm{tan}}^{–1}\mathrm{}\left(\mathrm{tan}\frac{\mathrm{\theta }}{\mathrm{2}}\right)\\ =\frac{\mathrm{\theta }}{\mathrm{2}}\end{array}$

Q.38

${\text{Prove that the sin}}^{-1}\mathrm{x}={\mathrm{cos}}^{-1}\mathrm{}\sqrt{1-{\mathrm{x}}^{2}}.$

Ans

$\begin{array}{l}\mathrm{Let}{\mathrm{sin}}^{-1}\mathrm{x}=\mathrm{\theta }.\dots ..\mathrm{}\left(1\right)\\ ⇒\mathrm{sin}\mathrm{\theta }=\mathrm{x}\\ ⇒{\mathrm{sin}}^{2}\mathrm{\theta }={\mathrm{x}}^{2}\\ ⇒1-{\mathrm{cos}}^{2}\mathrm{\theta }={\mathrm{x}}^{2}\\ ⇒{\mathrm{cos}}^{2}\mathrm{\theta }=1-{\mathrm{x}}^{2}\\ ⇒\mathrm{cos}\mathrm{\theta }=\sqrt{1-{\mathrm{x}}^{2}}\\ ⇒\mathrm{cos}\mathrm{\theta }=\sqrt{1-{\mathrm{x}}^{2}}\\ ⇒\mathrm{\theta }={\mathrm{cos}}^{-1}\sqrt{1-{\mathrm{x}}^{2}}.\dots .\left(2\right)\\ \mathrm{From}\left(1\right)\mathrm{and}\left(2\right)\mathrm{we}\mathrm{get},\\ {\mathrm{sin}}^{-1}\mathrm{x}={\mathrm{cos}}^{-1}\mathrm{}\sqrt{1-{\mathrm{x}}^{2}}\end{array}$

Q.39

${\text{Prove that cos}}^{-1}\mathrm{}\left(-\mathrm{x}\right)=\mathrm{\pi }-{\mathrm{cos}}^{-1}\left(\mathrm{x}\right).$

Ans

$\begin{array}{l}\mathrm{Let}{\mathrm{cos}}^{-1}\left(-\mathrm{x}\right)=\mathrm{\theta }.\dots .\left(1\right)\\ ⇒\mathrm{}-\mathrm{x}=\mathrm{cos}\mathrm{\theta }\\ ⇒\mathrm{x}=-\mathrm{cos}\mathrm{\theta }\\ ⇒\mathrm{x}=-\mathrm{cos}\left(\mathrm{\pi }-\mathrm{\theta }\right)\\ ⇒\mathrm{\pi }-\mathrm{\theta }={\mathrm{cos}}^{-1}\mathrm{x}\\ ⇒\mathrm{\theta }=\mathrm{}\mathrm{\pi }-{\mathrm{cos}}^{-1}\mathrm{x}.\dots \mathrm{}\left(2\right)\\ \mathrm{From}\left(1\right)\mathrm{and}\left(2\right)\mathrm{we}\mathrm{get},\\ ⇒{\mathrm{cos}}^{-1}\mathrm{}\left(-\mathrm{x}\right)\mathrm{}=\mathrm{\pi }-{\mathrm{cos}}^{-1}\mathrm{x}\end{array}$

Q.40

$\text{Evaluate the cos}\left({\mathrm{cosec}}^{-1}\mathrm{x}+{\mathrm{sec}}^{-1}\mathrm{x}\right).$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},\\ \mathrm{cos}\left({\mathrm{cosec}}^{-1}\mathrm{x}+{\mathrm{sec}}^{-1}\mathrm{x}\right)\\ =\mathrm{cos}\mathrm{}\left(\frac{\mathrm{\pi }}{2}\right)\left[âˆµ{\mathrm{cosec}}^{-1}\mathrm{x}+{\mathrm{sec}}^{-1}\mathrm{x}\frac{\mathrm{\pi }}{2}\right]\\ =\mathrm{}0\end{array}$

Q.41

${\text{Evaluate the tan}}^{-1}2\mathrm{}+{\mathrm{tan}}^{-1}3.$

Ans

$\begin{array}{l}{\text{We have, tan}}^{-1}2\text{}+{\text{tan}}^{-1}3{\text{=tan}}^{-1}\left(\frac{2+3}{1-2×3}\right)\\ \text{}={\text{tan}}^{-1}\left(\frac{5}{1-6}\right)\\ \text{}={\text{tan}}^{-1}\left(\frac{5}{-5}\right)\\ \text{}={\text{tan}}^{-1}\left(-1\right)\\ \text{}={\text{tan}}^{-1}\left(1\right)\\ \text{}=\text{}\frac{\pi }{4}\end{array}$

Q.42

${\text{Evaluate the tan}}^{-1}\frac{2}{11}\mathrm{}+{\mathrm{tan}}^{-1}\frac{7}{24}.$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},{\mathrm{tan}}^{-1}\frac{2}{11}+{\mathrm{tan}}^{-1}\frac{7}{24}\mathrm{}={\mathrm{tan}}^{-1}\left(\frac{\frac{2}{11}+\frac{7}{24}}{1-\frac{2}{11}×\frac{7}{24}}\right)\\ ={\mathrm{tan}}^{-1}\left(\frac{\frac{48+77}{11×24}}{1-\frac{14}{11×24}}\right)\\ ={\mathrm{tan}}^{-1}\left(\frac{125}{264-14}\right)\\ ={\mathrm{tan}}^{-1}\left(\frac{125}{250}\right)\\ ={\mathrm{tan}}^{-1}\left(\frac{1}{2}\right)\end{array}$

Q.43

$\text{Prove that}{\mathrm{sin}}^{–1}\mathrm{x}={\mathrm{cosec}}^{–1}\text{\hspace{0.17em}}\frac{1}{\mathrm{x}}.$

Ans

$\begin{array}{l}{\mathrm{Letsin}}^{-1}\mathrm{x}=\mathrm{\theta }.\dots \dots .\left(1\right)\\ ⇒\mathrm{sin}\mathrm{\theta }=\mathrm{x}\\ ⇒\mathrm{cosec}\mathrm{\theta }=\frac{1}{\mathrm{x}}\\ ⇒\mathrm{\theta }={\mathrm{cosec}}^{-1}=\left(\frac{1}{\mathrm{x}}\right)\mathrm{}.\dots \dots \left(2\right)\\ \mathrm{From}\left(1\right)\mathrm{and}\left(2\right)\mathrm{we}\mathrm{get},\\ {\mathrm{sin}}^{-1}\mathrm{x}={\mathrm{cosec}}^{-1}\left(\frac{1}{\mathrm{x}}\right)\end{array}$

Q.44

$\mathrm{Put}\mathrm{in}\mathrm{the}\mathrm{simplest}\mathrm{form},{\mathrm{tan}}^{–1}\frac{\sqrt{1+{\mathrm{x}}^{\mathrm{2}}}–1}{\mathrm{x}}\mathrm{.}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},{\mathrm{tan}}^{-1}\mathrm{}\left(\frac{\sqrt{1+{\mathrm{x}}^{2}-1}}{\mathrm{x}}\right)\\ \mathrm{Put}\mathrm{x}=\mathrm{tan\theta }\\ ={\mathrm{tan}}^{-1}\mathrm{}\left(\frac{\sqrt{1+{\mathrm{tan}}^{2}\mathrm{\theta }-1}}{\mathrm{tan\theta }}\right)\\ ={\mathrm{tan}}^{-1}\mathrm{}\left(\frac{\sqrt{{\mathrm{sec}}^{2}\mathrm{\theta }-1}}{\mathrm{tan\theta }}\right)\\ ={\mathrm{tan}}^{-1}\mathrm{}\left(\frac{\mathrm{sec\theta }-1}{\mathrm{tan\theta }}\right)\\ ={\mathrm{tan}}^{-1}\mathrm{}\left(\frac{\frac{1-\mathrm{cos\theta }}{\mathrm{cos\theta }}}{\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}}\right)\\ ={\mathrm{tan}}^{-1}\mathrm{}\left(\frac{1-\mathrm{cos\theta }}{\mathrm{sin\theta }}\right)\\ ={\mathrm{tan}}^{-1}\mathrm{}\left(\frac{2{\mathrm{sin}}^{2}\frac{\mathrm{\theta }}{2}}{2\mathrm{sin}\frac{\mathrm{\theta }}{2}\mathrm{cos}\frac{\mathrm{\theta }}{2}}\right)\\ ={\mathrm{tan}}^{-1}\mathrm{}\left(\frac{\mathrm{sin}\frac{\mathrm{\theta }}{2}}{\mathrm{cos}\frac{\mathrm{\theta }}{2}}\right)\\ ={\mathrm{tan}}^{-1}\mathrm{}\left(\mathrm{tan}\frac{\mathrm{\theta }}{2}\right)\\ =\mathrm{}\frac{\mathrm{\theta }}{2}\mathrm{}\left[âˆµ\mathrm{}\mathrm{tan\theta }\mathrm{}=\mathrm{}\mathrm{x}\mathrm{}⇒\mathrm{}\mathrm{\theta }\mathrm{}=\mathrm{}{\mathrm{tan}}^{-1}\mathrm{x}\right]\\ =\mathrm{}\frac{1}{2}\mathrm{}{\mathrm{tan}}^{-1}\mathrm{x}\end{array}$

Q.45

${\text{Prove that sin}}^{–1}\mathrm{x}+{\mathrm{cos}}^{–1}\mathrm{x}=\frac{\mathrm{\pi }}{\mathrm{2}}.$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have}\mathrm{to}\mathrm{prove}\mathrm{that}\\ {\mathrm{sin}}^{-1}\mathrm{x}\mathrm{}+\mathrm{}{\mathrm{cos}}^{-1}\mathrm{x}\mathrm{}=\mathrm{}\frac{\mathrm{\pi }}{2}.\\ \mathrm{Let}{\mathrm{sin}}^{-1}\mathrm{x}\mathrm{}=\mathrm{}\mathrm{\theta }.\dots ..\left(1\right)\\ ⇒\mathrm{}\mathrm{sin}\mathrm{ }\mathrm{\theta }\mathrm{}=\mathrm{}\mathrm{x}\\ ⇒\mathrm{}\mathrm{cos}\left(\frac{\mathrm{\pi }}{2}-\mathrm{\theta }\right)\mathrm{}=\mathrm{}\mathrm{x}\mathrm{}\left[âˆµ\mathrm{}\mathrm{sin\theta }\mathrm{}=\mathrm{}\mathrm{cos}\mathrm{}\left(\frac{\mathrm{\pi }}{2}-\mathrm{\theta }\right)\right]\\ ⇒\mathrm{}\frac{\mathrm{\pi }}{2}-\mathrm{\theta }\mathrm{}=\mathrm{}{\mathrm{cos}}^{-1}\mathrm{x}\\ ⇒\mathrm{}{\mathrm{cos}}^{-1}\mathrm{x}\mathrm{}=\mathrm{}\frac{\mathrm{\pi }}{2}-\mathrm{}\mathrm{\theta }\\ ⇒\mathrm{}\mathrm{\theta }\mathrm{}+\mathrm{}{\mathrm{cos}}^{-1}\mathrm{x}\mathrm{}=\mathrm{}\frac{\mathrm{\pi }}{2}\\ ⇒\mathrm{}{\mathrm{sin}}^{-1}\mathrm{x}+\mathrm{}{\mathrm{cos}}^{-1}\mathrm{x}\mathrm{}=\mathrm{}\frac{\mathrm{\pi }}{2}\mathrm{}\left(\mathrm{using}\mathrm{}\left(1\right)\right)\end{array}$

Q.46

$\mathrm{Show}\mathrm{that}{\mathrm{tan}}^{–1}\frac{\mathrm{2}}{\mathrm{11}}+{\mathrm{tan}}^{–1}\frac{\mathrm{7}}{\mathrm{24}}+{\mathrm{tan}}^{–1}\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{\pi }}{\mathrm{4}}\mathrm{.}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},\\ {\mathrm{tan}}^{–1}\frac{\mathrm{2}}{\mathrm{11}}+{\mathrm{tan}}^{–1}\frac{\mathrm{7}}{\mathrm{24}}+{\mathrm{tan}}^{–1}\frac{\mathrm{1}}{\mathrm{3}}\\ =\mathrm{}{\mathrm{tan}}^{-1}\mathrm{}\left(\frac{\frac{2}{11}+\frac{7}{24}}{1-\frac{2}{11}×\frac{7}{24}}\right)\mathrm{}+\mathrm{}{\mathrm{tan}}^{-1}\frac{1}{3}\\ =\mathrm{}{\mathrm{tan}}^{-1}\mathrm{}\left(\frac{\frac{48+77}{11×24}}{1-\frac{14}{11×24}}\right)\mathrm{}+\mathrm{}{\mathrm{tan}}^{-1}\frac{1}{3}\\ =\mathrm{}{\mathrm{tan}}^{-1}\mathrm{}\left(\frac{125}{264-14}\right)\mathrm{}+\mathrm{}{\mathrm{tan}}^{-1}\frac{1}{3}\\ =\mathrm{}{\mathrm{tan}}^{-1}\mathrm{}\left(\frac{125}{250}\right)\mathrm{}+\mathrm{}{\mathrm{tan}}^{-1}\frac{1}{3}\\ =\mathrm{}{\mathrm{tan}}^{-1}\mathrm{}\frac{1}{2}+{\mathrm{tan}}^{-1}\frac{1}{3}\\ =\mathrm{}{\mathrm{tan}}^{-1}\mathrm{}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}×\frac{1}{3}}\right)\\ =\mathrm{}{\mathrm{tan}}^{-1}\mathrm{}\left(\frac{\frac{5}{6}}{\frac{5}{6}}\right)\\ =\mathrm{}{\mathrm{tan}}^{-1}\mathrm{}\left(1\right)\\ =\mathrm{}\frac{\mathrm{\pi }}{4}\end{array}$

Q.47

$\begin{array}{l}\mathrm{If}\mathrm{two}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{are}{\mathrm{tan}}^{–1}2\mathrm{and}{\mathrm{tan}}^{–1}3,\\ \mathrm{than}\mathrm{find}\mathrm{the}\mathrm{third}\mathrm{angle}.\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{third}\mathrm{angle}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{be}\mathrm{\theta },\mathrm{then}\mathrm{we}\mathrm{have}\\ \mathrm{\theta }+{\mathrm{tan}}^{-1}+\mathrm{}{\mathrm{tan}}^{-1}\mathrm{}3\mathrm{}=\mathrm{}\mathrm{\pi }\mathrm{}\left[âˆµ\mathrm{}\mathrm{A}+\mathrm{B}+\mathrm{C}\mathrm{}=\mathrm{}\mathrm{\pi }\right]\\ ⇒\mathrm{}\mathrm{\theta }\mathrm{}+\mathrm{}{\mathrm{tan}}^{-1}\mathrm{}\left(\frac{2+3}{1-2×3}\right)\mathrm{}=\mathrm{}\mathrm{\pi }\\ ⇒\mathrm{}\mathrm{\theta }\mathrm{}+\mathrm{}{\mathrm{tan}}^{-1}\mathrm{}\left(\frac{5}{1-6}\right)\mathrm{}=\mathrm{}\mathrm{\pi }\\ ⇒\mathrm{}\mathrm{\theta }\mathrm{}+\mathrm{}{\mathrm{tan}}^{-1}\mathrm{}\left(\frac{5}{-5}\right)\mathrm{}=\mathrm{}\mathrm{\pi }\\ ⇒\mathrm{}\mathrm{\theta }\mathrm{}+\mathrm{}{\mathrm{tan}}^{-1}\mathrm{}\left(1\right)\mathrm{}=\mathrm{}\mathrm{\pi }\\ \mathrm{}⇒\mathrm{}\mathrm{\theta }\mathrm{}+\mathrm{}\frac{3\mathrm{\pi }}{4}\mathrm{}=\mathrm{}\mathrm{\pi }\mathrm{}\left[âˆµ\mathrm{}{\mathrm{tan}}^{-1}\left(-1\right)\mathrm{}=\mathrm{}\frac{3\mathrm{\pi }}{4}\right]\\ ⇒\mathrm{}\mathrm{\theta }\mathrm{}=\mathrm{}\mathrm{\pi }-\mathrm{}\frac{3\mathrm{\pi }}{4}\\ ⇒\mathrm{}\mathrm{\theta }\mathrm{}+\mathrm{}\frac{\mathrm{\pi }}{4}\mathrm{}\\ \mathrm{Therefore},\mathrm{the}\mathrm{third}\mathrm{angle}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{is}\frac{\mathrm{\pi }}{4}.\end{array}$

Q.48

$\mathrm{Solve}\mathrm{for}\mathrm{x},{\mathrm{sin}}^{–1}\left(1–\mathrm{x}\right)–2{\mathrm{sin}}^{–1}\mathrm{x}=\frac{\mathrm{\pi }}{\mathrm{2}}\mathrm{.}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},\\ {\mathrm{sin}}^{-1}\left(1-\mathrm{x}\right)-2{\mathrm{sin}}^{-1}\mathrm{x}\mathrm{}=\frac{\mathrm{\pi }}{2}\\ ⇒\mathrm{}{\mathrm{sin}}^{-1}\mathrm{}\left(1-\mathrm{x}\right)\mathrm{}=\mathrm{}\frac{\mathrm{\pi }}{2}\mathrm{}+\mathrm{}2\mathrm{}{\mathrm{sin}}^{-1}\mathrm{x}\\ ⇒\mathrm{}\left(1-\mathrm{x}\right)\mathrm{}=\mathrm{}\mathrm{sin}\mathrm{}\left(\frac{\mathrm{\pi }}{2}+2{\mathrm{sin}}^{-1}\mathrm{x}\right)\\ ⇒\mathrm{}\left(1-\mathrm{x}\right)\mathrm{}=\mathrm{}\mathrm{cos}\mathrm{}\left(2\mathrm{}{\mathrm{sin}}^{-1}\mathrm{x}\right)\\ ⇒\mathrm{}{\mathrm{cos}}^{-1}\left(1-\mathrm{x}\right)\mathrm{}=\mathrm{}2{\mathrm{sin}}^{-1}\mathrm{x}\\ ⇒\mathrm{}{\mathrm{sin}}^{-1\mathrm{}}\sqrt{1-{\left(1-\mathrm{x}\right)}^{2}}=\mathrm{}{\mathrm{sin}}^{-1}\left(2\mathrm{x}\sqrt{1-{\mathrm{x}}^{2}}\right)\\ ⇒\sqrt{1-{\left(1-\mathrm{x}\right)}^{2}}\mathrm{}=\mathrm{}\left(2\mathrm{x}\mathrm{}\sqrt{1-{\mathrm{x}}^{2}}\right)\\ \mathrm{On}\mathrm{squaring}\mathrm{both}\mathrm{sides},\\ ⇒1–{\left(1-\mathrm{x}\right)}^{2}=4{\mathrm{x}}^{2}\left(1-{\mathrm{x}}^{2}\right)\\ ⇒\mathrm{}1-1-{\mathrm{x}}^{2}+2\mathrm{x}=\mathrm{}4{\mathrm{x}}^{2}-4{\mathrm{x}}^{4}\\ ⇒\mathrm{}4{\mathrm{x}}^{4}-5{\mathrm{x}}^{2}+2\mathrm{x}\mathrm{}=\mathrm{}0\\ ⇒\mathrm{}\mathrm{x}\mathrm{}\left(4{\mathrm{x}}^{3}-5\mathrm{x}+2\right)\mathrm{}=\mathrm{}0\\ ⇒\mathrm{}\mathrm{x}=\mathrm{}0\mathrm{}\mathrm{or}\\ 4{\mathrm{x}}^{3}-5\mathrm{x}+2=\mathrm{}0\\ \mathrm{Put}\mathrm{x}=...-1, 0, 1, 2,.\dots \mathrm{we}\mathrm{get}\mathrm{no}\mathrm{real}\mathrm{vlue}\mathrm{of}\mathrm{x}\\ \mathrm{satisfy}\mathrm{above}\mathrm{cubic}\mathrm{equation}\mathrm{in}\left[-1,\mathrm{}1\right]\\ \mathrm{Hence}\mathrm{x}= 0\mathrm{is}\mathrm{only}\mathrm{solution}.\end{array}$

Q.49

$\mathrm{Solve}\mathrm{for}\mathrm{x},{\mathrm{tan}}^{–1}2\mathrm{x}+{\mathrm{tan}}^{–1}3\mathrm{x}=\frac{\mathrm{\pi }}{\mathrm{4}}\mathrm{.}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},{\mathrm{tan}}^{-1}\mathrm{}2\mathrm{x}+\mathrm{}{\mathrm{tan}}^{-1}\mathrm{}2\mathrm{x}\mathrm{}=\mathrm{}\frac{\mathrm{\pi }}{4}\\ ⇒\mathrm{}{\mathrm{tan}}^{-1}\left(\frac{2\mathrm{x}+3\mathrm{x}}{1-2\mathrm{x}\mathrm{}×\mathrm{}3\mathrm{x}}\right)\mathrm{}=\mathrm{}\frac{\mathrm{\pi }}{4}\\ ⇒\mathrm{}{\mathrm{tan}}^{-1}\left(\frac{5\mathrm{x}}{1-\mathrm{}6{\mathrm{x}}^{2}}\right)\mathrm{}=\mathrm{}\frac{\mathrm{\pi }}{4}\\ ⇒\mathrm{}\frac{5\mathrm{x}}{1-\mathrm{}6{\mathrm{x}}^{2}}\mathrm{}=\mathrm{}\mathrm{tan}\mathrm{}\frac{\mathrm{\pi }}{4}\\ ⇒\mathrm{}\frac{5\mathrm{x}}{1-\mathrm{}6{\mathrm{x}}^{2}}\mathrm{}=\mathrm{}1\\ ⇒\mathrm{}5\mathrm{x}\mathrm{}=\mathrm{}1-\mathrm{}6{\mathrm{x}}^{2}\\ ⇒\mathrm{}6{\mathrm{x}}^{2}\mathrm{}=5\mathrm{x}-\mathrm{}1\mathrm{}=\mathrm{}0\\ ⇒\mathrm{}\left(6\mathrm{x}-1\right)\left(\mathrm{x}+1\right)\mathrm{}=\mathrm{}0\\ ⇒\mathrm{}\mathrm{x}\mathrm{}=\mathrm{}-\mathrm{}\left(\mathrm{rejected}\mathrm{as}\mathrm{does}\mathrm{not}\mathrm{satisfy}\mathrm{equation}\right)\\ ⇒\mathrm{}\mathrm{x}\mathrm{}=\mathrm{}\frac{1}{6}\mathrm{is}\mathrm{the}\mathrm{only}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{equation}.\end{array}$

Q.50

$\mathrm{Put}\mathrm{in}\mathrm{the}\mathrm{simplest}\mathrm{form},{\mathrm{tan}}^{–1}\mathrm{ }\left(\frac{\mathrm{cos}\mathrm{\theta }}{1+\mathrm{sin}\mathrm{\theta }}\right)\mathrm{.}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},{\mathrm{tan}}^{–1}\left(\frac{\mathrm{cos}\mathrm{\theta }}{1+\mathrm{sin}\mathrm{\theta }}\right)\\ \mathrm{}={\mathrm{tan}}^{–1}\left(\frac{{\mathrm{cos}}^{2}\mathrm{}\frac{\mathrm{\theta }}{2}-{\mathrm{sin}}^{2}\frac{\mathrm{\theta }}{2}}{{\left(\mathrm{cos}\frac{\mathrm{\theta }}{2}-\mathrm{sin}\frac{\mathrm{\theta }}{2}\right)}^{2}}\right)\\ ={\mathrm{tan}}^{-1}\mathrm{}\left(\frac{\left(\mathrm{cos}\frac{\mathrm{\theta }}{2}+\mathrm{sin}\frac{\mathrm{\theta }}{2}\right)\left(\mathrm{cos}\frac{\mathrm{\theta }}{2}-\mathrm{sin}\frac{\mathrm{\theta }}{2}\right)}{\left(\mathrm{cos}\frac{\mathrm{\theta }}{2}-\mathrm{sin}\frac{\mathrm{\theta }}{2}\right)}\right)\\ ={\mathrm{tan}}^{-1}\mathrm{}\left(\frac{\left(\mathrm{cos}\frac{\mathrm{\theta }}{2}\mathrm{}-\mathrm{}\mathrm{sin}\frac{\mathrm{\theta }}{2}\right)}{\left(\mathrm{cos}\frac{\mathrm{\theta }}{2}+\mathrm{sin}\frac{\mathrm{\theta }}{2}\right)}\right)\\ \mathrm{Divide}{\mathrm{N}}^{1}\mathrm{and}{\mathrm{D}}^{1}\mathrm{by}\mathrm{cos}\frac{\mathrm{\theta }}{2}\\ ={\mathrm{tan}}^{–1}\frac{\left(1–\mathrm{tan}\frac{\mathrm{\theta }}{2}\right)}{\left(1+\mathrm{tan}\frac{\mathrm{\theta }}{2}\right)}\\ ={\mathrm{tan}}^{-1}\mathrm{}\frac{\left(\mathrm{tan}\mathrm{}\frac{\mathrm{\pi }}{4}-\mathrm{tan}\frac{\mathrm{\theta }}{2}\right)}{\left(1+\mathrm{tan}\mathrm{}\frac{\mathrm{\pi }}{4}.\mathrm{tan}\frac{\mathrm{\theta }}{2}\right)}\\ ={\mathrm{tan}}^{–1}\left[\left(\mathrm{tan}\frac{\mathrm{\pi }}{4}-\frac{\mathrm{\theta }}{2}\right)\right]\\ =\mathrm{}\frac{\mathrm{\pi }}{4}-\frac{\mathrm{\theta }}{2}\end{array}$

Q.51

$\mathrm{Show}\mathrm{that}{\mathrm{tan}}^{–1}\frac{\mathrm{1}}{5}+{\mathrm{tan}}^{–1}\frac{\mathrm{1}}{7}+{\mathrm{tan}}^{–1}\frac{\mathrm{1}}{3}+{\mathrm{tan}}^{–1}\frac{\mathrm{1}}{\mathrm{8}}\mathrm{}=\mathrm{}\frac{\mathrm{\pi }}{4}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},\\ {\mathrm{tan}}^{–1}\frac{\mathrm{1}}{5}+{\mathrm{tan}}^{–1}\frac{\mathrm{1}}{7}+{\mathrm{tan}}^{–1}\frac{\mathrm{1}}{3}+{\mathrm{tan}}^{–1}\frac{\mathrm{1}}{\mathrm{8}}\\ ={\mathrm{tan}}^{-1}\left(\frac{\frac{1}{5}+\frac{1}{7}}{1-\frac{1}{5}×\frac{1}{7}}\right)+{\mathrm{tan}}^{-1}\left(\frac{\frac{1}{3}+\frac{1}{8}}{1-\frac{1}{3}×\frac{1}{8}}\right)\\ =\mathrm{}{\mathrm{tan}}^{-1}\left(\frac{\frac{7+5}{35}}{1-\frac{1}{35}}\right)+{\mathrm{tan}}^{-1}\left(\frac{\frac{8+3}{24}}{1-\frac{1}{24}}\right)\\ ={\mathrm{tan}}^{-1}\left(\frac{\frac{12}{35}}{\frac{34}{35}}\right)+{\mathrm{tan}}^{-1}\left(\frac{\frac{11}{24}}{\frac{23}{24}}\right)\\ =\mathrm{}{\mathrm{tan}}^{-1}\left(\frac{12}{34}\right)+{\mathrm{tan}}^{-1}\left(\frac{11}{23}\right)\\ =\mathrm{}{\mathrm{tan}}^{-1}\left(\frac{6}{17}\right)+{\mathrm{tan}}^{-1}\left(\frac{11}{23}\right)\\ ={\mathrm{tan}}^{-1}\left(\frac{\frac{6}{17}+\frac{11}{23}}{1-\frac{6}{17}×\frac{11}{23}}\right)\\ =\mathrm{}{\mathrm{tan}}^{-1}\left(\frac{\frac{138+187}{17×23}}{\frac{391-66}{17×23}}\right)\\ =\mathrm{}{\mathrm{tan}}^{-1}\left(\frac{325}{325}\right)\\ =\mathrm{}{\mathrm{tan}}^{-1}\left(1\right)\\ =\frac{\mathrm{\pi }}{4}\\ =\mathrm{}1\end{array}$

Q.52

$\mathrm{Put}\mathrm{in}\mathrm{the}\mathrm{simplest}\mathrm{form},{\mathrm{tan}}^{–1}\left(\frac{\mathrm{cos}\mathrm{x}}{1–\mathrm{sin}\mathrm{x}}\right)\mathrm{.}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},{\mathrm{tan}}^{–1}\left(\frac{\mathrm{cos}\mathrm{x}}{1–\mathrm{sin}\mathrm{x}}\right)\\ \mathrm{}={\mathrm{tan}}^{–1}\left(\frac{{\mathrm{cos}}^{2}\mathrm{}\frac{\mathrm{x}}{2}-{\mathrm{sin}}^{2}\frac{\mathrm{x}}{2}}{{\left(\mathrm{cos}\frac{\mathrm{x}}{2}-\mathrm{sin}\frac{\mathrm{x}}{2}\right)}^{2}}\right)\\ {\mathrm{tan}}^{-1}=\left(\frac{\left(\mathrm{cos}\frac{\mathrm{x}}{2}+\mathrm{sin}\frac{\mathrm{x}}{2}\right)\left(\mathrm{cos}\frac{\mathrm{x}}{2}-\mathrm{sin}\frac{\mathrm{x}}{2}\right)}{\left(\mathrm{cos}\frac{\mathrm{x}}{2}-\mathrm{sin}\frac{\mathrm{x}}{2}\right)}\right)\\ {\mathrm{tan}}^{-1}=\left(\frac{\left(\mathrm{cos}\frac{\mathrm{x}}{2}+\mathrm{sin}\frac{\mathrm{x}}{2}\right)}{\left(\mathrm{cos}\frac{\mathrm{x}}{2}-\mathrm{sin}\frac{\mathrm{x}}{2}\right)}\right)\\ \mathrm{Divide}{\mathrm{N}}^{1}\mathrm{and}{\mathrm{D}}^{1}\mathrm{by}\mathrm{cos}\frac{\mathrm{x}}{2}\\ ={\mathrm{tan}}^{–1}\frac{\left(1+\mathrm{tan}\frac{\mathrm{x}}{2}\right)}{\left(1–\mathrm{tan}\frac{\mathrm{x}}{2}\right)}\\ ={\mathrm{tan}}^{-1}\mathrm{}\frac{\left(\mathrm{tan}\mathrm{}\frac{\mathrm{\pi }}{4}+\mathrm{tan}\frac{\mathrm{x}}{2}\right)}{\left(\mathrm{1}-\mathrm{tan}\mathrm{}\frac{\mathrm{\pi }}{4}.\mathrm{tan}\frac{\mathrm{x}}{2}\right)}\\ ={\mathrm{tan}}^{–1}\left[\left(\mathrm{tan}\frac{\mathrm{\pi }}{4}+\frac{\mathrm{x}}{2}\right)\right]\\ =\mathrm{}\frac{\mathrm{\pi }}{4}+\frac{\mathrm{x}}{2}\end{array}$

Q.53

${\text{Show that tan}}^{-1}\frac{2}{11}+{\mathrm{tan}}^{-1}\frac{7}{24}+{\mathrm{tan}}^{-1}\frac{3}{4}\mathrm{}={\mathrm{tan}}^{-1}\mathrm{}2.$

Ans

$\begin{array}{l}\mathrm{LHS}={\mathrm{tan}}^{-1}\frac{2}{11}+{\mathrm{tan}}^{-1}\frac{7}{24}+{\mathrm{tan}}^{-1}\frac{3}{4}\mathrm{}\\ ={\mathrm{tan}}^{-1}\left(\frac{\frac{2}{11}+\frac{7}{24}}{1-\frac{2}{11}×\frac{7}{24}}\right)\mathrm{}+\mathrm{}{\mathrm{tan}}^{-1}\frac{3}{4}\\ ={\mathrm{tan}}^{-1}\left(\frac{\frac{48+77}{11×24}}{1-\frac{14}{11×24}}\right)\mathrm{}+\mathrm{}{\mathrm{tan}}^{-1}\frac{3}{4}\\ ={\mathrm{tan}}^{-1}\left(\frac{125}{264-14}\right)\mathrm{}+\mathrm{}{\mathrm{tan}}^{-1}\frac{3}{4}\\ ={\mathrm{tan}}^{-1}\left(\frac{125}{250}\right)\mathrm{}+\mathrm{}{\mathrm{tan}}^{-1}\frac{3}{4}\\ ={\mathrm{tan}}^{-1}\left(\frac{1}{2}\right)\mathrm{}+\mathrm{}{\mathrm{tan}}^{-1}\frac{3}{4}\\ ={\mathrm{tan}}^{-1}\left(\frac{\frac{1}{2}+\frac{3}{4}}{1-\frac{1}{2}×\frac{3}{4}}\right)\mathrm{}\\ ={\mathrm{tan}}^{-1}\left(\frac{\frac{2+3}{4}}{\frac{8-3}{8}}\right)\mathrm{}\\ ={\mathrm{tan}}^{-1}\left(\frac{\frac{5}{4}}{\frac{5}{8}}\right)\mathrm{}\\ ={\mathrm{tan}}^{-1}\left(\frac{5}{4}×\frac{8}{5}\right)\mathrm{}\\ ={\mathrm{tan}}^{-1}2\mathrm{}\left(\mathrm{RHS}\mathrm{Proved}\right)\mathrm{}\end{array}$

Q.54

$\mathrm{Evaluate}\mathrm{cot}\left({\mathrm{tan}}^{–1}\frac{\mathrm{1}}{\mathrm{3}}+{\mathrm{tan}}^{–1}\frac{\mathrm{1}}{\mathrm{5}}+{\mathrm{tan}}^{–1}\frac{\mathrm{1}}{\mathrm{7}}+{\mathrm{tan}}^{–1}\frac{\mathrm{1}}{\mathrm{8}}\right)$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},\\ \mathrm{cot}\left[{\mathrm{tan}}^{–1}\frac{\mathrm{1}}{\mathrm{3}}+{\mathrm{tan}}^{–1}\frac{\mathrm{1}}{\mathrm{5}}+{\mathrm{tan}}^{–1}\frac{\mathrm{1}}{\mathrm{7}}+{\mathrm{tan}}^{–1}\frac{\mathrm{1}}{\mathrm{8}}\right]\\ =\mathrm{cot}\left[{\mathrm{tan}}^{–1}\frac{\mathrm{1}}{5}+{\mathrm{tan}}^{–1}\frac{\mathrm{1}}{7}+{\mathrm{tan}}^{–1}\frac{\mathrm{1}}{3}+{\mathrm{tan}}^{–1}\frac{\mathrm{1}}{\mathrm{8}}\right]\\ =\mathrm{}\mathrm{cot}\mathrm{}\left[{\mathrm{tan}}^{-1}\left(\frac{\frac{1}{5}+\frac{1}{7}}{1-\frac{1}{5}×\frac{1}{7}}\right)+{\mathrm{tan}}^{-1}\left(\frac{\frac{1}{3}+\frac{1}{8}}{1-\frac{1}{3}×\frac{1}{8}}\right)\right]\\ =\mathrm{}\mathrm{cot}\mathrm{}\left[{\mathrm{tan}}^{-1}\left(\frac{\frac{7+5}{35}}{1-\frac{1}{35}}\right)+{\mathrm{tan}}^{-1}\left(\frac{\frac{8+3}{24}}{1-\frac{1}{24}}\right)\right]\\ =\mathrm{}\mathrm{cot}\mathrm{}\left[{\mathrm{tan}}^{-1}\left(\frac{\frac{12}{35}}{\frac{34}{35}}\right)+{\mathrm{tan}}^{-1}\left(\frac{\frac{11}{24}}{\frac{23}{24}}\right)\right]\\ =\mathrm{}\mathrm{cot}\mathrm{}\left[{\mathrm{tan}}^{-1}\left(\frac{12}{34}\right)+{\mathrm{tan}}^{-1}\left(\frac{11}{23}\right)\right]\\ =\mathrm{}\mathrm{cot}\mathrm{}\left[{\mathrm{tan}}^{-1}\left(\frac{6}{17}\right)+{\mathrm{tan}}^{-1}\left(\frac{11}{23}\right)\right]\\ =\mathrm{}\mathrm{cot}\mathrm{}+\left[{\mathrm{tan}}^{-1}\left(\frac{\frac{138+187}{17×23}}{\frac{391-66}{17×23}}\right)\right]\\ =\mathrm{}\mathrm{cot}\mathrm{}+\left[{\mathrm{tan}}^{-1}\left(\frac{325}{325}\right)\right]\\ \\ =\mathrm{}\mathrm{cot}\mathrm{}+\left[{\mathrm{tan}}^{-1}\left(1\right)\right]\\ =\mathrm{}\mathrm{cot}\mathrm{}\frac{\mathrm{\pi }}{4}\\ =\mathrm{}1\end{array}$

Q.55

$\begin{array}{l}\mathrm{If}{\mathrm{sin}}^{-1}\mathrm{x}+{\mathrm{sin}}^{-1}\mathrm{y}+{\mathrm{sin}}^{-1}\mathrm{z}\mathrm{}=\mathrm{}\frac{\mathrm{\pi }}{2},\mathrm{then}\\ \mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}+2\mathrm{xyz}.\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},\\ {\mathrm{sin}}^{-1}\mathrm{x}+{\mathrm{sin}}^{-1}\mathrm{y}\mathrm{}+\mathrm{}{\mathrm{sin}}^{-1}\mathrm{z}\mathrm{}=\mathrm{}\frac{\mathrm{\pi }}{2}\\ ⇒{\mathrm{cos}}^{-1}\sqrt{1-{\mathrm{x}}^{2}}+{\mathrm{cos}}^{-1}\sqrt{1-{\mathrm{y}}^{2}}\mathrm{}=\frac{\mathrm{\pi }}{2}{\mathrm{sin}}^{-1}\mathrm{z}\\ ⇒{\mathrm{cos}}^{-1}\left[\sqrt{1-{\mathrm{x}}^{2}}\sqrt{1-{\mathrm{y}}^{2}}-\sqrt{\left\{1-\left(1-{\mathrm{x}}^{2}\right)\right\}}\mathrm{}\sqrt{\left\{1-\left(1-{\mathrm{y}}^{2}\right)\right\}}\right]\mathrm{}=\mathrm{}{\mathrm{cos}}^{-1}\mathrm{z}\\ ⇒\mathrm{}\sqrt{1-{\mathrm{x}}^{2}}\sqrt{1-{\mathrm{y}}^{2}}-\sqrt{\left({\mathrm{x}}^{2}\right)}\sqrt{\left({\mathrm{y}}^{2}\right)}\mathrm{}=\mathrm{}\mathrm{z}\\ ⇒\mathrm{}\sqrt{1-{\mathrm{x}}^{2}}\sqrt{1-{\mathrm{y}}^{2}}=\mathrm{xy}\mathrm{}+\mathrm{}\mathrm{z}\\ ⇒\mathrm{}\sqrt{1-{\mathrm{x}}^{2}}\sqrt{1-{\mathrm{y}}^{2}}=\mathrm{xy}\mathrm{}+\mathrm{}\mathrm{z}\\ \mathrm{Squaring}\mathrm{both}\mathrm{sides}\\ {\left(\mathrm{xy}+\mathrm{z}\right)}^{2}\mathrm{}=\mathrm{}\left(1-{\mathrm{x}}^{2}\right)\left(1-{\mathrm{y}}^{2}\right)\\ ⇒\mathrm{}{\mathrm{x}}^{2}{\mathrm{y}}^{2}+{\mathrm{z}}^{2}+2\mathrm{xyz}\mathrm{}=\mathrm{}1-{\mathrm{x}}^{2}-{\mathrm{y}}^{2}+{\mathrm{x}}^{2}{\mathrm{y}}^{2}\\ ⇒\mathrm{}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}+2\mathrm{xyz}\mathrm{}=\mathrm{}1\\ \mathrm{Hence},\mathrm{value}\mathrm{of}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}+2\mathrm{xyz}\mathrm{}\mathrm{is}1.\end{array}$

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### 1. Which chapter’s notes can I get for Class 12 mathematics on Extramarks?

Extramarks, an online learning platform, provides informative and detailed notes along with all topics and sub-topics of all the chapters for Mathematics Class 12. Students are advised to study with the help of these academic notes.

Chapter 1: Relations and Functions

Chapter 2: Inverse Trigonometric Functions

Chapter 3: Matrices

Chapter 4: Determinants

Chapter 5: Continuity and Differentiability

Chapter 6: Applications of Derivatives

Chapter 7: Integrals

Chapter 8: Application of Integrals

Chapter 9: Differential Equations

Chapter 10: Vector Algebra

Chapter 11: Three Dimensional Geometry

Chapter 12: Linear programming

Chapter 13: Probability

### 2. Will using the Class 12 Mathematics Chapter 2 notes help me in my exam preparation?

Yes, definitely. The chapter 2 mathematics class 12 notes include every minute detail that the students must know in order to attain high scores. It includes solutions to all-important questions along with certain key points, formulae, derivations and more that will enhance the preparation level. Some tips, tricks and short-cut methods will also help students to tackle difficult questions in no time.