Important Questions For CBSE Class 12 Maths Chapter 2 Inverse Trigonometric Functions

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CBSE Class 12 Maths Chapter 2 Important Questions: Key Concepts and Solutions

Overview of Chapter 2

Chapter 2 of CBSE Class 12 Maths focuses on Inverse Trigonometric Functions. Understanding these functions is crucial for solving various mathematical problems effectively.

CBSE Class 12 Maths Chapter 2 Inverse Trigonometric Functions Important Questions

Question 1. The principal value of \(\cos ^{-1}\left(\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{\sqrt{2}}\right)\) is

  1. \(\frac{\pi}{12}\)
  2. \(\pi\)
  3. \(\frac{\pi}{3}\)
  4. \(\frac{\pi}{6}\)

Solution: 1. \(\frac{\pi}{12}\)

⇒ \(\cos ^{-1}\left(\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{\sqrt{2}}\right)=\frac{\pi}{3}+\left(-\frac{\pi}{4}\right)\)

= \(\frac{4 \pi-3 \pi}{12}=\frac{\pi}{12}\)

Read and Learn More CBSE Class 12 Maths Important Question and Answers

Question 2. The principal value of \(\tan ^{-1}\left(\tan \frac{9 \pi}{8}\right)\) is:

  1. \(\frac{\pi}{8}\)
  2. \(\frac{3 \pi}{8}\)
  3. \(-\frac{\pi}{8}\)
  4. \(-\frac{3 \pi}{8}\)

Solution: 1. \(\frac{\pi}{8}\)

⇒ \(\tan ^{-1}\left[\tan \left(\frac{9 \pi}{8}\right)\right]=\tan ^{-1}\left[\tan \left(\pi+\frac{\pi}{8}\right)\right]\)

= \(\tan ^{-1}\left[\tan \left(\frac{\pi}{8}\right)\right]=\frac{\pi}{8} \in\left(-\frac{\pi}{2} \cdot \frac{\pi}{2}\right)\)

CBSE Important Questions for Class 12 Maths

Question 3. What is the domain of the function cos-1(2x-3)?

  1. [-1, 1]
  2. (1,2)
  3. (-1, 1)
  4. [1,2]

Solution: 4. [1,2]

For the given function: -1 ≤ 2x – 3 ≤ 1

⇒ 2 ⇒ 2x ⇒ 4

⇒ 1 ≤ x ≤ 2

Question 4. The principal value of [tan-1 √3 – cot-1 (-√3)] is

  1. π
  2. \(-\frac{\pi}{2}\)
  3. 0
  4. \(2 \sqrt{3}\)

Solution: 2. \(-\frac{\pi}{2}\)

⇒ \(\tan ^{-1}(\sqrt{3})-\cot ^{-1}(-\sqrt{3})=\frac{\pi}{3}-\left(\pi-\cot ^{-1}(\sqrt{3})\right)\)

=\(\frac{\pi}{3}-\left(\pi-\frac{\pi}{6}\right)=\frac{\pi}{3}-\frac{5 \pi}{6}=\frac{-3 \pi}{6}=\frac{-\pi}{2}\)

Important Questions For CBSE Class 12 Maths Chapter 2

Question 5. The range or the principal valurbrance of the function y = sec-1 x is ____

or,

The principal value of \(\cos ^{-1}\left(-\frac{1}{2}\right)\) is

Solution:

y = \(\sec ^{-1} x\) (given)

Range of \(\sec ^{-1} x\) is [0, π]-{π/2}

Or

Let \(\cos ^{-1}\left(-\frac{1}{2}\right)=y\)

⇒ \(\cos y=-\frac{1}{2} \Rightarrow \cos y=\cos \left[\pi-\frac{\pi}{3}\right]\)

⇒ \(\cos y=\cos \frac{2 \pi}{3} \Rightarrow y=\frac{2 \pi}{3} \in[0, \pi]\)

CBSE Important Questions for Class 12 Maths

Question 6. Simplify \(\sec ^{-1}\left(\frac{1}{2 x^2-1}\right), 0<x<\frac{1}{\sqrt{2}} \text {. }\)

Solution:

Let y = \(\sec ^{-1}\left(\frac{1}{2 x^2-1}\right), 0<x<\frac{1}{\sqrt{2}}\)

Put x = \(\cos \theta\); then \(\theta=\cos ^{-1} x\)

y = \(\sec ^{-1}\left(\frac{1}{2 \cos ^2 \theta-1}\right)\)

= \(\sec ^{-1}\left(\frac{1}{\cos 2 \theta}\right)=\sec ^{-1}(\sec 2 \theta)=2 \theta=2 \cos ^{-1} x\)

Question 7. Prove that: \(3 \sin ^{-1} x=\sin ^{-1}\left(3 x-4 x^3\right), x \in\left[-\frac{1}{2}, \frac{1}{2}\right]\)

Solution:

To prove: \(3 \sin ^{-1} x=\sin ^{-1}\left(3 x-4 x^3\right), x \in\left[-\frac{1}{2}, \frac{1}{2}\right]\)

R.H.S = sin-1(3x – 4x³)

Put x = sinθ :  θ = sin-1x

⇒ R.H.S = sin-1 (3sinθ – 4sin³θ) – sin-1(sin3θ) = 3θ = 3sin-1x = L.H.S

Hence, proved.

CBSE Class 12 Maths Chapter 2 Conclusion

Focusing on these important questions will enhance your preparation for CBSE Class 12 Maths Chapter 2. Regular practice will build confidence and improve your problem-solving skills.

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