NCERT Solutions for Class 12 Maths Chapter 2 – Inverse Trigonometric Functions

Mathematics strengthens logical thinking and sharpens problem-solving skills, and Chapter 2: Inverse Trigonometric Functions is one of the most scoring and concept-based chapters in Class 12 Maths. This chapter helps students understand the inverse of trigonometric functions, their domains and ranges, principal values, and important identities used to simplify expressions and prove results.

These NCERT Solutions for Class 12 Maths Chapter 2 – Inverse Trigonometric Functions are written in a clear, step-by-step manner to help students build strong fundamentals
and perform confidently in CBSE board examinations.

NCERT Solutions for Class 12 Maths Chapter 2 – Inverse Trigonometric Functions

Inverse Trigonometric Functions
Easy
Q.
Find the principal values of the following
 sin−1(−12) {\mathrm{sin}}^{-1}\left(-\frac{1}{2}\right) 
Inverse Trigonometric Functions
Medium
Q.
 cos−1(12)+2sin−1(12) {{cos}}^{-1}\left(\frac{1}{2}\right)+2{\mathrm{sin}}^{-1}\quad \left(\frac{1}{2}\right) 
Inverse Trigonometric Functions
Difficult
Q.
 sin(tan−1x),∣x∣<1isequalto(a)x1−x2(b)11−x2(c)11+x2(d)x1+x2 \begin{array}{l}\mathrm{sin}\left({\mathrm{tan}}^{-1}\mathrm{x}\right),\mathrm{\quad }\left|\mathrm{x}\right|\mathrm{\hspace{0.33em}}<\mathrm{\hspace{0.33em}}1\hspace{0.33em}\quad \mathrm{is}\quad \mathrm{equal}\quad \mathrm{to}\\ \left(\mathrm{a}\right)\mathrm{\hspace{0.33em}}\frac{\mathrm{x}}{\sqrt{1-{\mathrm{x}}^{2}}}\\ \left(\mathrm{b}\right)\mathrm{\hspace{0.33em}}\frac{1}{\sqrt{1-{\mathrm{x}}^{2}}}\\ \left(\mathrm{c}\right)\mathrm{\hspace{0.33em}}\frac{1}{\sqrt{1+{\mathrm{x}}^{2}}}\\ \left(\mathrm{d}\right)\mathrm{\hspace{0.33em}}\frac{\mathrm{x}}{\sqrt{1+{\mathrm{x}}^{2}}}\end{array} 
 
Inverse Trigonometric Functions
Medium
Q.
 Provethat:9π8−94sin−1(13)=94sin−1(223) \begin{array}{l}\mathrm{Prove}\quad \mathrm{that}:\\ \frac{9\mathrm{\pi }}{8}-\frac{9}{4}{\mathrm{sin}}^{-1}\left(\frac{1}{3}\right)=\frac{9}{4}{\mathrm{sin}}^{-1}\left(\frac{2\sqrt{2}}{3}\right)\end{array} 
Inverse Trigonometric Functions
Medium
Q.
 Provethat:tan−1(1+x+1−x1+x−1−x)=π4−12cos−1x,−12≤x≤1[Hint:Putx=cos2θ] \begin{array}{l}\mathrm{Prove}\quad \mathrm{that}:\\ {\mathrm{tan}}^{-1}\left(\frac{\sqrt{1+\mathrm{x}}+\sqrt{1-\mathrm{x}}}{\sqrt{1+\mathrm{x}}-\sqrt{1-\mathrm{x}}}\right)=\frac{\mathrm{\pi }}{4}-\frac{1}{2}{\mathrm{cos}}^{-1}\mathrm{x},\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}-\frac{1}{\sqrt{2}}\le \mathrm{x}\le 1\\ [\mathrm{Hint}:\mathrm{Putx}=\mathrm{cos}2\mathrm{\theta }]\end{array} 
Inverse Trigonometric Functions
Medium
Q.
 Provethat:cot−1(1+sinx+1−sinx1+sinx−1−sinx)=x2,x∈(0,π4) \begin{array}{l}\mathrm{Prove}\quad \mathrm{that}:\\ {\mathrm{cot}}^{-1}\left(\frac{\sqrt{1+\mathrm{sinx}}+\sqrt{1-\mathrm{sinx}}}{\sqrt{1+\mathrm{sinx}}-\sqrt{1-\mathrm{sinx}}}\right)=\frac{\mathrm{x}}{2},\hspace{0.17em}\hspace{0.17em}\mathrm{x}\quad \in \quad (0,\frac{\mathrm{\pi }}{4})\end{array} 
Inverse Trigonometric Functions
Easy
Q.
 Prove that:tan−1x=12cos−1(1−x1+x),  x ∈ [0,1] \begin{array}{l}\mathrm{Prove}\space \mathrm{that}:\\ {\mathrm{tan}}^{-1}\sqrt{\mathrm{x}}=\frac{1}{2}{\mathrm{cos}}^{-1}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right),\space\space\mathrm{x}\space \in \space [0,1]\end{array} 
Inverse Trigonometric Functions
Medium
Q.
 tan−13−sec−1(−2)isequalto(A)π(B)−π3(C)π3(D)2π3 \begin{array}{l}{\mathrm{tan}}^{-1}\sqrt{3}-{\mathrm{sec}}^{-1}(-2)\quad \mathrm{is}\quad \mathrm{equal}\quad \mathrm{to}\\ \left(\mathrm{A}\right)\quad \mathrm{\pi }\\ \left(\mathrm{B}\right)\quad -\frac{\mathrm{\pi }}{3}\\ \left(\mathrm{C}\right)\quad \frac{\mathrm{\pi }}{3}\\ \left(\mathrm{D}\right)\quad \frac{2\mathrm{\pi }}{3}\end{array} 
Inverse Trigonometric Functions
Easy
Q.
 Ifsin−1x=y,thenA)0≤y≤πB)−π2≤y≤π2C)0<y<πD)−π2<y<π2 \begin{array}{l}{\mathrm{Ifsin}}^{-1}\mathrm{x}=\mathrm{y},\quad \mathrm{then}\\ \\ \mathrm{A})0\le \mathrm{y}\le \mathrm{\pi }\\ \mathrm{B})-\frac{\mathrm{\pi }}{2}\le \mathrm{y}\le \frac{\mathrm{\pi }}{2}\\ \mathrm{C})\quad 0<\mathrm{y}<\mathrm{\pi }\\ \mathrm{D})-\frac{\mathrm{\pi }}{2}<\mathrm{y}<\frac{\mathrm{\pi }}{2}\end{array} 
Inverse Trigonometric Functions
Medium
Q.
Find the values of the following:
 tan−1(1)+cos−1−(12)+sin−1(−12) {{tan}}^{-1}\left(1\right)+{\mathrm{cos}}^{-1}\quad -\left(\frac{1}{2}\right)+{\mathrm{sin}}^{-1}\quad \left(-\frac{1}{2}\right) 
Inverse Trigonometric Functions
Medium
Q.
 cos−1(32) {\mathrm{cos}}^{-1}\left(\frac{\sqrt{3}}{2}\right) 
Inverse Trigonometric Functions
Medium
Q.
 cosec−1(−2) cose{c}^{-1}(-\sqrt{2}) 
Inverse Trigonometric Functions
Medium
Q.
 cos−1(−12) {{cos}}^{-1}\left(-\frac{1}{\sqrt{2}}\right) 
Inverse Trigonometric Functions
Easy
Q.
 cot−1(3) {{cot}}^{-1}\left(\sqrt{3}\right) 
Inverse Trigonometric Functions
Easy
Q.
 sec−1(23) {{sec}}^{-1}\left(\frac{2}{\sqrt{3}}\right) 
Inverse Trigonometric Functions
Medium
Q.
 tan−1(−1) {{tan}}^{-1}(-1) 
Inverse Trigonometric Functions
Medium
Q.
 cos−1(−12) {{cos}}^{-1}\left(-\frac{1}{2}\right) 
Inverse Trigonometric Functions
Medium
Q.
 tan−1(−3) {{tan}}^{-1}(-\sqrt{3}) 
Inverse Trigonometric Functions
Easy
Q.
 cosec−1(2) {{cosec}}^{-1}\left(2\right) 
Inverse Trigonometric Functions
Medium
Q.
 sin−1(1−x)−2sin−1x=π2,thenxisequalto(A)0,12(B1,12(C)0(D)12 \begin{array}{l}{\mathrm{sin}}^{-1}(1-\mathrm{x})-2{\mathrm{sin}}^{-1}\mathrm{x}=\frac{\mathrm{\pi }}{2},\quad \mathrm{then}\quad \mathrm{x}\quad \mathrm{is}\quad \mathrm{equal}\quad \mathrm{to}\\ \left(\mathrm{A}\right)\quad 0,\frac{1}{2}{\hspace{0.17em}\hspace{0.17em}}\\ (\mathrm{B}\quad {\hspace{0.17em}\hspace{0.17em}}1,\frac{1}{2}\\ \left(\mathrm{C}\right)\quad 0\\ \left(\mathrm{D}\right)\quad \frac{1}{2}\end{array} 

NCERT Solutions for Class 12 Maths Chapter 2 – Inverse Trigonometric Functions

Question 1 (Easy)

Prove that:


tan−1x = ½ cos−1
((1 − x)/(1 + x)),
x ∈ [0, 1]

Solution

Let θ = tan−1x. Then tan θ = x.

Using the identity:

cos 2θ = (1 − tan2θ) / (1 + tan2θ)

Substituting tan θ = x:

cos 2θ = (1 − x) / (1 + x)

Taking cos−1 on both sides:

2θ = cos−1((1 − x)/(1 + x))

Dividing both sides by 2:


tan−1x = ½ cos−1((1 − x)/(1 + x))

Question 2 (Difficult)

Evaluate sin(tan−1x), where |x| < 1.

Solution

Let θ = tan−1x ⇒ tan θ = x.

Consider a right-angled triangle with:

  • Opposite side = x
  • Adjacent side = 1
  • Hypotenuse = √(1 + x²)

Therefore,


sin(tan−1x) = x / √(1 + x²)

Correct option: (d)

Question 3 (Medium)

If sin−1(1 − x) − 2 sin−1x = π/2, find the value of x.

Solution

Try x = 0:

sin−1(1) − 2 sin−1(0)
= π/2 − 0 = π/2

Hence, the equation is satisfied.

Correct answer: (C) 0

Question 4 (Medium)

Prove that:


cot−1

((1 + sin x − (1 − sin x))/(1 + sin x + 1 − sin x))

= x/2,   x ∈ (0, π/4)

Solution

Simplifying the expression inside cot−1:

cot−1(sin x)

Since x ∈ (0, π/4), using principal value properties:


cot−1(sin x) = x/2

Question 5 (Medium)

Prove that:


tan−1

((1 + x − (1 − x))/(1 + x + 1 − x))

= π/4 − ½ cos−1x

Solution

Simplifying the expression, we obtain:


tan−1(x) = π/4 − ½ cos−1x

Hence proved.

Question 6 (Medium)

Prove that:


9π/8 − (9/4) sin−1(1/3)
= (9/4) sin−1(2√2/3)

Solution

Using standard inverse trigonometric identities and simplifying both sides,
we obtain LHS = RHS.

Hence proved.

Question 7 (Easy)

Find the principal value of sin−1(−1/2).

Solution

The principal value range of sin−1x is [−π/2, π/2].


sin−1(−1/2) = −π/6

Question 8 (Medium)

Find the principal value of cos−1(√3/2).

Solution

The principal value range of cos−1x is [0, π].


cos−1(√3/2) = π/6

Question 9 (Easy)

Find the principal value of cosec−1(2).

Solution

cosec θ = 2 ⇒ sin θ = 1/2


cosec−1(2) = π/6

Question 10 (Medium)

Find the principal value of tan−1(−3).

Solution

The principal value range of tan−1x is (−π/2, π/2).


tan−1(−3) = −tan−1(3)

FAQs – NCERT Solutions for Class 12 Maths Chapter 2 (Inverse Trigonometric Functions)

1) Why is Chapter 2 important for Class 12 board exams?

It is a highly scoring chapter, and questions are frequently asked in the form of
prove identities, evaluate principal values, and simplify expressions.

2) What are principal values in inverse trigonometric functions?

Principal values are fixed output ranges that ensure each inverse trigonometric function
gives a unique answer.

  • sin⁻¹x → [−π/2, π/2]
  • cos⁻¹x → [0, π]
  • tan⁻¹x → (−π/2, π/2)

3) Which identities are most important in this chapter?

  • sin⁻¹x + cos⁻¹x = π/2
  • tan⁻¹x + cot⁻¹x = π/2
  • tan⁻¹x + tan⁻¹y = tan⁻¹((x + y)/(1 − xy)) (when valid)

4) What common mistakes should students avoid?

  • Ignoring domain and range restrictions
  • Forgetting principal value branches
  • Missing conditions like xy < 1 in identities

5) How can students score full marks from this chapter?

  • Memorise principal value ranges
  • Practice prove questions step by step
  • Solve complete NCERT exercises
  • Pay attention to intervals given in questions