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. 

Students should stay tuned with Extramarks to update themselves with all the latest information regarding CBSE exams, syllabus, paper patterns, and more. 

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)

graph of y= 〖sin〗^(-1) (x)

2. y= cos-1(x)

graph of y= 〖cos〗^(-1) (x)

  1. y = tan-1(x)

3. y = tan-1(x)

  1. y= cot-1(x)

4. y= cot-1(x)

  1. y = cosec-1(x)

y = cosec-1(x)

  1. y= sec-1(x)

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 ]

Additional Formulas: 

  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

If sin−135=x, find the value of cos x.

Ans

Here, sin−135=x⇒sinx=35Now, cos x = 1−sin2x=1−352=1625=45

Q.2

Solvetan−12x+tan−13x=π4

Ans

tan−12x+tan−13x=π4⇒tan−12x+3x1−2x×3x=π4∴tan−1x+tan−1y=tan−1x+y1−xy⇒tan−15x1−6x2=π4∴5x1−6x2=tanπ4⇒5x1−6x2=1or5x=1−6x2⇒6x2+5x−1=0i.e. 6x−1x+1=0Which gives, x = 16,x=−1Since x = -1 does not satisfy the equation as the L.H.S. of the equationbecomes negative, x = 16 is the only solution of the given equation.

Q.3

Prove that tan−11+x−1−x1+x+1−x=π4−12cos−1x

Ans

Here, L.H.S. = tam−11+x−1−x1+x+ 1−xLet x = cos2θ⇒θ=12cos−1xThenL.H.S.= tan−11+cos2θ−1−cos2θ1+cos2θ+1−cos2θ= tan−12cos2θ−2sin2θ2cos2θ+2sin2θ= tan−12cosθ−2sinθ2cosθ+2sinθ= tan−1cosθ−sinθcosθ+sinθ= tan−11−tanθ1+tanθ= tan−1tanπ4−tanθ1+tanπ4tanθ= tan−1tan−1π4−θ=π4−θ=π4−12cos−1x=R.H.SProved

Q.4

Prove that cot−11+sinx+1−sinx1+sinx−1−sinx=x2,×∈0,π4.

Ans

L.H.S.=cot−11+sinx+1−sinx1+sinx−1−sinx=cot−11+sinx+1−sinx1+sinx−1−sinx×1+sinx+1+sinx1+sinx+1−sinx=cot−11+sinx+1−sinx21+sinx−1+sinx=cot−11+sinx+1−sinx+21−sin2x2sinx=cot−121+cosx2sinx=cot−12cos2x22sinx2cos x2=cot−1cotx2=x2=R.H.S

Q.5

Show that sin−11213+cos−145+tan−16316=π

Ans

sin−11213=x,cos−145=ytan−16316=z⇒sinx=1213,cosy=45andtanz6316∴cosx=513, siny=34,tanx=125andtany34Now, tanx+y=tanx+tany1−tan×tany=125+341−125×34=−6316⇒tanx+y=−tanz⇒tanx+y=tanπ−z∴x+y=π−z⇒x+y+z=π

Q.6

Solve the following equation:tan−1x−1x−2+tan−1x+1x+2=π4

Ans

tan−1x−1x−2+tan−1x+1x+2=π4⇒tan−1x−1x−2+x+1x+21−x−1x−2×x+1x+2=π4⇒x2+x−2+x2−x−2x2−4−x2+1=1⇒2x2−4−3=1⇒2x2−4=−3⇒x2=12⇒x=±12

Q.7

Evaluatetan12sin−12x1+x2+cos−11−y21+y2,x<1,y>0 and xy > 1

Ans

Let x = tan θ⇒θtan−1xand=tan−1y then,tan12sin−12x1+x2+cos−11−y21+y2= tan12sin−12tanθ1+tan2θ+cos−11−tan21+tan2=tan12sin−1sin2θ+cos−1cos2=tan122θ+2=tanθ+

Q.8

Show that sin−135−sin−1817=cos−18485

Ans

Let sin−135 =x and sin−1817=y⇒sin−135 =sinx and 817=sinyas cosx = 1−sin−2x=1−925=45and cos y = 1−sin2y=1−64289=1517Now, cosx−y=cosxcosy+sinxsiny=45×1517+35×817cosx−y=8485⇒x−y=cos−18485

Q.9

Evaluate tan–11+cos–1–12+sin–1–12

Ans

tan−11+cos–1–12+sin–1–12tan−1tanπ4+cos–1−cos π3+sin–1−sinπ6=π4+cos–1cosπ− π3+sin–1sin−π6 we know range of tan−1→−π2, π2,cos−1→0,πandsin−1→−π2, π2∴tan−11+cos−1−12+sin−1−12 =π4+2π3−π6=3π+8π−2π12=9π12=3π4

Q.10

Write the function tan−1cosx−sinxcosx+sinxin the simplest form.

Ans

tan−1cosx−sinxcosx+sinx=tan−1cosxcosx−sinxcosxcosxcosx+sinxcosx=tan−11−tanx1+tanx=tan−1tanπ4−tanx1+tanπ4tanx=tan−1tanπ4−x∴tanA−B= tanA–tanB1+tanAtanB=π4−x

Q.11

Express tan-1cos x1-sin x,π2< x<π2in the simplest form.

Ans

tan–1cos x1–sin x=tan–1sinπ2−x1−cosπ2−x=tan−1sinπ−2x21−cosπ−2x2=tan−12sinπ−2x42sin2π−2x4=tan−1cosπ−2x4sinπ−2x4=tan−1cotπ−2x4=tan−1tanπ2−π−2x4=tan−1tanπ4+π2=π4+x2

Q.12

Find the value of sin–1sin2π3

Ans

The range of inverse function is −π2,−π2.Now, sin−1sin2π3=sin−1sinπ−π3=sin−1sinπ3=π3

Q.13

Write tan−11x2−1,x>1 in the simplest form.

Ans

Let x = cosec θ⇒ θ = cosec−1xThen, tan−11x2−1=tan−11cosec2−1=tan−11cotθ=tan−1tanθ=θ=cosec−1x

Q.14

Prove that tan−1x+tan−12x1−x2=tan−13x−x31−3x2

Ans

Let x = tan θNow, L.H.S. = tan−1x+tan−12x1−x2=tan−1tanθ+tan−12tanθ1−tan2θ=θ+tan−1tan2θ=θ+2θ=3θ.…..1R.H.S.= tan−13tanθ−tan3θ1−3tan2θ= tan−13tanθ=3θ.….2by 1and2L.H.S.=R.H.S. proved

Q.15

Show that: tan−112+tan−1211=tan−134

Ans

L.H.S.= tan−112+2111−12×211tan−1x+tan−1y=tan−1x+y1−xy=tan−11520=tan−134= R.H.S.

Q.16

Find the principal value ofisin-1-12iicot-1-13

Ans

isin–1–12=y⇒siny=−12⇒siny=−sinπ6⇒siny=−sin−π6ory=−π6We know that the range of inverse sine function is −π2,π2∴sin−1−π6=−π6iiLet cot−1−13=y⇒coty=−13⇒coty=−13⇒coty=cotπ−π3⇒y=2π3We know that range of inverse contangent function 0,π∴cot−1−13=2π3

Q.17

Is the following statement true ?cos–1x=cosx–1

Ans

No, cos–1x¹≠cos x–1=1cos x

Q.18

Find the principal value of cos–132

Ans

Letcos–132=y⇒cosy=32Range of cos−1is0,πcosπ6=32⇒cos−132=π6

Q.19

If sin−1x=y, than what will be the range of y ?

Ans

principal value of sin−1 lies between −π2,π2−π2≤π2y≤π2

Q.20

Show that sin−12x1−x2=2sin−1x

Ans

sin−12x1−x2=2sin−1xlet x = sinθsin−12x1−x2=sin−12sinθ1−sin2θ=sin−12sinθcosθ=sin−1sin2θ=2θ=2sin−1x

Q.21

Find the value of cos−1cos13π6.

Ans

cos−1cos13π6=13π6But 13π6∉0,πcos13π6=cos2π+π6=cosπ6∴cos−1cos13π6=π6

Q.22

Find the value of tan–1211+tan–1724.

Ans

tan–1211+tan–1724.=tan–1211+7241−211×724tan−1x+tan−1y=tan−1x+y1−xy=tan−1125250=tan−112

Q.23

Find the value of tan–13a2x−x3a3−3ax2.

Ans

Letx=a tanθThen tan–13a2x−x3a3−3ax2=tan–13a2tanθ−a3tan3θa3−3a2tan2θ                          =tan–13tanθ−tan3θ1−3tan2θ                          =tan–1tan3θ                          =3θ=3tan–1xa

Q.24

Write 2 tan–1x in terms of sin–1x, cos–1x and tan–1x.

Ans

2 tan–1x= sin–12x1+x2 =cos–11−x21+x2 = tan–12x1−x2

Q.25

Write the domain and range of tan-1x.

Ans

Domain of tan–1x=RRange of ten–1x=−π2,π2

Q.26

Fill in the following blanks:

a)cos−1−x=.…b)cos−1 1x=.….

Ans

a)cos−1−x=π−cos−1xb)cos−1 1x=sec−1x

Q.27

Find the value of sin π3−sin−1−12.

Ans

sin π3−sin−1−12=sin π3+sin−1−12=sin π3+π6=sinπ2=1.

Q.28

Evaluatetan–1tan4π3.

Ans

The rage of inverse tangent function is −π2,π2.Now, tan−1tan4π3=tan−1tanπ+π3=tan−1π3=π3

Q.29

Solve sin 2cos−1cot2tan−1x=0.

Ans

Given equation is sin 2cos−1cot2tan−1x=0⇒ 2cos−1cot2tan−1x=nπ,where n is any integer⇒ cos−1cot2tan−1x=nπ/2⇒ cos−1cot2tan−1x=0,π2,πsince cos−1xliesin0,π⇒cot2tan−1x=cos0,cosπ2,cosπ=1,0,−1⇒cottan−12x/1−x2=1,0,−1⇒1/tantan−12x/1−x2=1,0,−1⇒1−x2/2x=1,0,−1⇒1−x2/2x=1,1−x2/2x=0 and 1−x2/2x=−1⇒x=−1±2,x=±1,x=1±2.

Q.30

Prove that sin−11213+cos−145=sinh−16365.

Ans

sin−11213=xandcos−145=y∴cosx513andsiny=35Now, sinx+y=sinxcosy+cosxsiny=1213,45+513,35=6365⇒x+y=sin−16365so,sin−11213+cos−145=sin−16365

Q.31

Prove that tan–115+tan–116 = tan–11129

Ans

Prove that tan–115+tan–116 = tan–11129L.H.S.= tan−115+tan−116=tan−115+161−15×16=tan−16+530−1= tan−11129=R.H.S. Proved.

Q.32

Find the principal value of sin−112.

Ans

Let sin−112=θ⇒sinθ=12⇒sinθ=sin π6⇒sinθ=sin π6∵The range of sin–−1xis π2,π2⇒θ=π6∴The principal value of sin−112isπ6.

Q.33

Final the principal value of sin−112.

Ans

Let sin−112=θ⇒sinθ=12⇒cosθ=−cos π4⇒sinθ=sin π4∵The range of sin–−1xis π2,π2⇒θ=π4∴The principal value of sin−112isπ4.

Q.34

Findthe principal value of cos–112.

Ans

Let cos−1−12=0⇒cosθ=−12⇒cosθ=−cos π4⇒cosθ=cos π4∵The range of cos–−1xis 0,π⇒cosθ=cos 3π4⇒θ=−3π4∴The principal value of cos−1−12is3π4.

Q.35

Evaluate the cot tan–1a+cot–1a.

Ans

We have,=cottan–1a+cot–1a=cotπ2∵tan–1a+cot–1a=π2=0

Q.36

Evaluate the sin sin-1x+cos-1x.

Ans

We have,=sinsin–1x+cos–1x=sinπ2∵sin–1x+cos–1x=π2=1

Q.37

Put in the simplest form: tan–11– cosθsinθ.

Ans

Wehave, tan–11–cos θsin θ=tan–12 sin2 θ22 sinθ2cosθ2=tan–1sin θ2cosθ2=tan–1tanθ2= θ2

Q.38

Prove that the sin−1x= cos−11−x2.

Ans

Let sin−1x=θ.…..1⇒sin θ=x⇒sin2θ =x2⇒1−cos2θ=x2⇒cos2θ=1−x2⇒cos θ=1−x2⇒cos θ=1−x2⇒θ = cos−11−x2.….2From 1and 2 we get,sin−1x= cos−11−x2

Q.39

Prove that cos−1 −x= π−cos−1x.

Ans

Let cos−1−x=θ .….1⇒−x = cos θ⇒x = −cos θ⇒x = −cos π −θ⇒π−θ = cos−1x ⇒ θ=π− cos−1x.…2From 1and 2 we get,⇒ cos−1−x=π−cos−1x

Q.40

Evaluate the cos cosec−1x+ sec−1x.

Ans

We have,cos cosec−1x+ sec−1x=cos π2∵ cosec−1x+sec−1xπ2=0

Q.41

Evaluate the tan−12+ tan−13.

Ans

We have, tan−12+tan−13 =tan−12+31−2×3 =tan−151−6 =tan−15−5 =tan−1−1 =tan−11 =π4

Q.42

Evaluate the tan−1211+ tan−1724.

Ans

We have, tan−1211+ tan−1724=tan−1211+7241−211×724 =tan−148+7711×241−1411×24 =tan−1125264−14 =tan−1125250 =tan−112

Q.43

Prove that sin–1x=cosec–1 1x.

Ans

Letsin−1x = θ.…….1⇒sin θ = x⇒ cosec θ=1x⇒θ=cosec−1=1x.……2From 1 and 2 we get,sin−1x= cosec−11x

Q.44

Put in the simplest form, tan–11+x2–1x.

Ans

We have, tan−11+x2−1xPutx = tanθ= tan−11+tan2θ−1tanθ= tan−1sec2θ−1tanθ= tan−1secθ−1tanθ= tan−11−cosθcosθsinθcosθ= tan−11−cosθsinθ= tan−12sin2θ22sinθ2cosθ2= tan−1sinθ2cosθ2= tan−1tanθ2=θ2∵tanθ=x⇒θ=tan−1x=12tan−1x

Q.45

Prove that sin–1x+cos–1x=π2.

Ans

We have to prove thatsin−1x+cos−1x=π2.Let sin−1x=θ.…..1⇒sin θ=x⇒cosπ2−θ=x∵sinθ=cosπ2−θ⇒π2−θ=cos−1x⇒cos−1x=π2−θ⇒θ+cos−1x=π2⇒sin−1x+cos−1x=π2using1

Q.46

Show that tan–1211+tan–1724+tan–113=π4.

Ans

We have,tan–1211+tan–1724+tan–113=tan−1211+7241−211×724+tan−113=tan−148+7711×241−1411×24+tan−113=tan−1125264−14+tan−113=tan−1125250+tan−113=tan−112+tan−113=tan−112+131−12×13=tan−15656=tan−11=π4

Q.47

If two angles of a triangle are tan–12andtan–1 3,than find the third angle.

Ans

Let third angle of a triangle be θ, then we haveθ+tan−1+tan−13=π∵A+B+C=π⇒θ+tan−12+31−2×3=π⇒θ+tan−151−6=π⇒θ+tan−15−5=π⇒θ+tan−11=π⇒θ+3π4=π∵tan−1−1=3π4⇒θ=π−3π4⇒θ+π4Therefore, the third angle of a triangle is π4.

Q.48

Solve for x, sin–11–x–2sin–1x=π2.

Ans

We have,sin−11−x−2sin−1x=π2⇒sin−11−x=π2+2sin−1x⇒1−x=sinπ2+2sin−1x⇒1−x=cos2sin−1x⇒cos−11−x=2sin−1x⇒sin−11−1−x2=sin−12x1−x2⇒1−1−x2=2x1−x2On squaring both sides,⇒ 1–1−x2=4x21−x2⇒1−1−x2+2x=4x2−4x4⇒4x4−5x2+2x=0⇒x4x3−5x+2=0⇒x=0or4x3−5x+2=0 Put x = ... -1, 0, 1, 2, .… we get no real vlue of xsatisfy above cubic equation in −1,1Hence x = 0 is only solution.

Q.49

Solve for x, tan–12x+tan–13x=π4.

Ans

We have, tan−12x+tan−12x=π4⇒tan−12x+3x1−2x×3x=π4⇒tan−15x1−6x2=π4⇒5x1−6x2=tanπ4⇒5x1−6x2=1⇒5x=1−6x2⇒6x2=5x−1=0⇒6x−1x+1=0⇒x=−rejected as does not satisfy equation⇒x=16 is the only solution of the given equation.

Q.50

Put in the simplest form, tan–1 cos θ1 + sin θ.

Ans

We have, tan–1cosθ1+ sinθ=tan–1cos2θ2−sin2θ2cosθ2−sinθ22= tan−1cosθ2+sinθ2cosθ2−sinθ2cosθ2−sinθ2=tan−1cosθ2−sinθ2cosθ2+sinθ2Divide N1 and D1 by cosθ2= tan–11– tanθ21+ tanθ2= tan−1tanπ4−tanθ21+tanπ4.tanθ2= tan–1tanπ4−θ2=π4−θ2

Q.51

Show that tan–115+ tan–117+ tan–113+ tan–118=π4

Ans

We have,tan–115+ tan–117+ tan–113+ tan–118=tan−115+171−15×17+tan−113+181−13×18=tan−17+5351−135+tan−18+3241−124=tan−112353435+tan−111242324=tan−11234+tan−11123=tan−1617+tan−11123= tan−1617+11231−617×1123=tan−1138+18717×23391−6617×23=tan−1325325=tan−11=π4=1

Q.52

Put in the simplest form, tan–1cosx1–sinx.

Ans

We have, tan–1cosx1–sinx=tan–1cos2x2−sin2x2cosx2−sinx22tan−1=cosx2+sinx2cosx2−sinx2cosx2−sinx2tan−1=cosx2+sinx2cosx2−sinx2Divide N1 and D1 by cosx2= tan–11+ tanx21– tanx2= tan−1tanπ4+tanx21−tanπ4.tanx2= tan–1tanπ4+x2=π4+x2

Q.53

Show that tan−1211 + tan−1724+tan−134=tan−12.

Ans

LHS = tan−1211 + tan−1724+tan−134=tan−1211+7241−211×724+tan−134=tan−148+7711×241−1411×24+tan−134=tan−1125264−14+tan−134=tan−1125250+tan−134=tan−112+tan−134=tan−112+341−12×34=tan−12+348−38=tan−15458=tan−154×85=tan−12(RHS Proved)

Q.54

Evaluate cot tan–113+tan–115+tan–117+tan–118

Ans

We have,cot tan–113+ tan–115+ tan–117+ tan–118= cot tan–115+ tan–117+ tan–113+ tan–118=cottan−115+171−15×17+tan−113+181−13×18=cottan−17+5351−135+tan−18+3241−124=cottan−112353435+tan−111242324=cottan−11234+tan−11123=cottan−1617+tan−11123=cot+tan−1138+18717×23391−6617×23=cot+tan−1325325=cot+tan−11=cotπ4=1

Q.55

If sin−1x+sin−1y+sin−1z=π2, then find the value of x2+y2+z2+2xyz.

Ans

We have,sin−1x+sin−1y+sin−1z=π2⇒cos−11−x2+cos−11−y2=π2sin−1z⇒cos−11−x21−y2−1−1−x21−1−y2=cos−1z⇒1−x21−y2−x2y2=z⇒1−x21−y2=xy+z⇒1−x21−y2=xy+zSquaring both sidesxy+z2=1−x21−y2⇒x2y2+z2+2xyz=1−x2−y2+x2y2⇒x2+y2+z2+2xyz=1Hence, value of x2+y2+z2+2xyzis 1.

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FAQs (Frequently Asked Questions)

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

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