exercise 2.1 class 12 maths ncert solutions || class 12 maths ncert solutions chapter 2 ex 2.1 || ex 2.1 class 12 maths ncert solutions || inverse trigonometric functions class 12 ncert solutions (English Medium)
NCERT Solutions for Class 12 Maths – Chapter 2 Exercise 2.1 (Inverse Trigonometric Functions)
Exercise 2.1 of Class 12 Maths NCERT Solutions focuses on the fundamental concepts of Inverse Trigonometric Functions, which is a key chapter in the CBSE Class 12 Maths curriculum. This exercise introduces students to the principal value branches of inverse trigonometric functions such as sin⁻¹x, cos⁻¹x, tan⁻¹x, and their respective domains and ranges. The Class 12 Maths Exercise 2.1 NCERT Solutions provide step-by-step explanations to help students understand how to evaluate and simplify inverse trigonometric expressions. These solutions are designed as per the latest CBSE guidelines and are ideal for English Medium students. Studying from the Class 12 Maths NCERT Solutions Chapter 2 Exercise 2.1 helps build a strong foundation for solving higher-order problems. The Exercise 2.1 Class 12 Maths NCERT Solutions are an excellent resource for mastering this topic and scoring well in board exams as well as competitive exams like JEE.
inverse trigonometric functions class 12 ncert solutions || ex 2.1 class 12 maths ncert solutions || exercise 2.1 class 12 maths ncert solutions || class 12 maths ncert solutions chapter 2 ex 2.1
Exercise 2.1
1. Find the principal values of the following:
\( \operatorname{Sin}^{-1}\left(-\frac{1}{2}\right) \)
\( \operatorname{Sin}^{-1}\left(-\frac{1}{2}\right) \)
Answer
Let us take \( \sin ^{-1}\left(-\frac{1}{2}\right)=x \)
Therefore,
\(
\operatorname{Sin} x=-\frac{1}{2}=-\sin \frac{\pi}{6}=\sin \left(-\frac{\pi}{6}\right)
\)
We know that principle value range of \( \sin ^{-1} \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)
And,
\(
\operatorname{Sin}\left(-\frac{\pi}{6}\right)=-\frac{1}{2}
\)
Therefore principle value of \( \sin ^{-1} \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)
2. Find the principal values of the following:
\( \operatorname{Cos}^{-1}\left(\frac{\sqrt{3}}{2}\right) \)
\( \operatorname{Cos}^{-1}\left(\frac{\sqrt{3}}{2}\right) \)
Answer
Let us take \( \cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)=x \)
Then, \( \cos x=\frac{\sqrt{3}}{2}=\cos \left(\frac{\pi}{6}\right) \)
We know that principle value range of \( \cos ^{-1} \) is \( [0, \pi] \)
And \( \cos \left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2} \)
Therefore, principle value of \( \cos ^{-1}\left(\frac{\sqrt{3}}{2}\right) \) is \( \frac{\pi}{6} \)
inverse trigonometric functions class 12 ncert solutions || ex 2.1 class 12 maths ncert solutions || exercise 2.1 class 12 maths ncert solutions || class 12 maths ncert solutions chapter 2 ex 2.1
3. Find the principal values of the following:
\( \operatorname{cosec}^{-1}(2) \)
\( \operatorname{cosec}^{-1}(2) \)
Answer
Let \( \operatorname{cosec}^{-1} 2=x \)
Therefore, \( \operatorname{cosec} x=2=\operatorname{cosec}\left(\frac{\pi}{6}\right) \)
And \( \operatorname{cosec}\left(\frac{\pi}{6}\right)=2 \)
We know that principle value range of \( \operatorname{cosec}^{-1} \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\} \)
Therefore, principle value of \( \operatorname{cosec}^{-1}(2) \) is \( \frac{\pi}{6} \).
4. Find the principal values of the following:
\( \tan ^{-1}(-\sqrt{3}) \)
\( \tan ^{-1}(-\sqrt{3}) \)
Answer
Let us take \( \tan ^{-1}(-\sqrt{3})=x \)
Then we get,
\( \tan x=-\sqrt{3}=-\tan \frac{\pi}{3}=\tan \left(-\frac{\pi}{3}\right) \)
And \( \tan \left(-\frac{\pi}{3}\right)=-\sqrt{3} \)
We know that principle value range of \( \tan -1 \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{3}\right] \)
Therefore, principle value of \( \tan ^{-1}(-\sqrt{3}) \) is \( \left(-\frac{\pi}{3}\right) \).
5. Find the principal values of the following:
\( \operatorname{Cos}^{-1}\left(-\frac{1}{2}\right) \)
\( \operatorname{Cos}^{-1}\left(-\frac{1}{2}\right) \)
Answer
Let us take \( \cos ^{-1}\left(-\frac{1}{2}\right)=x \)
Then we will get,
\( \operatorname{Cos} x=-\frac{1}{2}=-\cos \left(\frac{\pi}{3}\right)=\cos \left(\pi-\frac{\pi}{3}\right)=\cos \left(\frac{2 \pi}{3}\right) \)
And \( \cos \left(\frac{2 \pi}{3}\right)=-\frac{1}{2} \)
We know that principle value range of \( \cos ^{-1} \) is \( [0, \pi] \)
Therefore, principle value of \( \cos ^{-1}\left(-\frac{1}{2}\right) \) is \( \frac{2 \pi}{3} \).
inverse trigonometric functions class 12 ncert solutions || ex 2.1 class 12 maths ncert solutions || exercise 2.1 class 12 maths ncert solutions || class 12 maths ncert solutions chapter 2 ex 2.1
6. Find the principal values of the following:
\( \tan ^{-1}(-1) \)
\( \tan ^{-1}(-1) \)
Answer
Let us take \( \tan ^{-1}(-1)=x \) then we get, \( =\tan x=-1=-\tan \left(\frac{\pi}{4}\right) \)
\( =\tan \left(-\frac{\pi}{4}\right) \)
And \( \tan \left(-\frac{\pi}{4}\right)=-1 \)
We know that principle value range of \( \tan ^{-1} \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)
Therefore, principle value of \( \tan ^{-1}(-1) \) is \( -\frac{\pi}{4} \)
7. Find the principal values of the following:
\( \operatorname{Sec}^{-1}\left(\frac{2}{\sqrt{3}}\right) \)
\( \operatorname{Sec}^{-1}\left(\frac{2}{\sqrt{3}}\right) \)
Answer
Let \( \sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)=x \)
Then,
\( \operatorname{Sec} x=\frac{2}{\sqrt{3}}=\sec \left(\frac{\pi}{6}\right) \)
We know that range of the principle value branch of \( \sec ^{-1} \) is \( [0, \pi]-\left\{\frac{\pi}{2}\right\} \)
And \( \sec \left(\frac{\pi}{6}\right)=\frac{2}{\sqrt{3}} \)
Therefore principle value of \( \sec ^{-1}\left(\frac{2}{\sqrt{3}}\right) \) is \( \frac{\pi}{6} \)
8. Find the principal values of the following:
\( \operatorname{Cot}^{-1}(-\sqrt{3})=x \)
\( \operatorname{Cot}^{-1}(-\sqrt{3})=x \)
Answer
Let us consider \( \cot ^{-1}(-\sqrt{3})=x \)
Then we get,
\( \operatorname{Cot} x=-\sqrt{3}=-\cot \left(\frac{\pi}{6}\right)=\cot \left(\pi-\frac{\pi}{6}\right)=\cot \frac{5 \pi}{6} \)
We know that the range of the principal value branch of \( \cot -1 \) is \( [0, \pi] \).
And, \( \cot \left(\frac{5 \pi}{6}\right)=-\sqrt{3} \)
Therefore, the principle value of \( \cot ^{-1}(-\sqrt{3}) \) is \( \frac{5 \pi}{6} \).
9. Find the principal values of the following:
\( \operatorname{Cos}^{-1}\left(-\frac{1}{\sqrt{2}}\right) \)
\( \operatorname{Cos}^{-1}\left(-\frac{1}{\sqrt{2}}\right) \)
Answer
Let \( \cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)=x \)
Therefore,
\( \operatorname{Cos} x=-\frac{1}{\sqrt{2}}=-\cos \left(\frac{\pi}{4}\right)=\cos \left(\pi-\frac{\pi}{4}\right)=\cos \left(\frac{3 \pi}{4}\right) \)
We know that range of the principle value branch of \( \cos ^{-1} \) is \( [0, \pi] \)
And \( \cos \left(\frac{3 \pi}{4}\right)=-\frac{1}{\sqrt{2}} \)
Therefore principle value of \( \cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right) \) is \( \frac{3 \pi}{4} \)
10. Find the principal values of the following:
\( \operatorname{Cosec}^{-1}(-\sqrt{2}) \)
\( \operatorname{Cosec}^{-1}(-\sqrt{2}) \)
Answer
Let us take the values of \( \operatorname{cosec}^{-1}(-\sqrt{2}),=x \)
Then,
\( \operatorname{Cosec} x=-\sqrt{2}=\operatorname{cosec}\left(-\frac{\pi}{4}\right) \)
We know that range of the principle value branch of \( \operatorname{cosec}-1 \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\} \)
And \( \operatorname{cosec}\left(-\frac{\pi}{4}\right)=-\sqrt{2} \)
Therefore principle value of \( \operatorname{cosec}^{-1}(-\sqrt{2}) \) is \( -\frac{\pi}{4} \).
11. Find the values of the following:
\(
\tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)
\)
\(
\tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)
\)
Answer
Let us consider \( \tan ^{-1}(1)=x \) then we get
\( \tan x=1=\tan \frac{\pi}{4} \)
We know that range of the principle value branch of \( \tan ^{-1} \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \) Therefore, \( \tan ^{-1}(1)=\frac{\pi}{4} \)
Let \( \cos ^{-1}\left(-\frac{1}{2}\right)=y \)
\( \operatorname{Cos} y=-\frac{1}{2}=\cos \left(\pi-\frac{\pi}{3}\right)=\cos \left(\frac{2 \pi}{3}\right) \)
We know that range of the principle value branch of \( \cos ^{-1} \) is \( [0, \pi] \)
Therefore, \( \cos ^{-1}\left(-\frac{1}{2}\right)=\frac{2 \pi}{3} \)
Let \( \sin ^{-1}\left(-\frac{1}{2}\right)=z \)
\( \operatorname{Sin} z=-\sin \frac{\pi}{6}=\sin \left(-\frac{\pi}{6}\right) \)
We know that range of the principle value branch of \( \sin ^{-1} \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)
Therefore \( \sin ^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6} \)
Now,
\( \tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin \left(-\frac{1}{2}\right) \)
\( =\frac{\pi}{4}+\frac{2 \pi}{3}-\frac{\pi}{6}=\frac{3 \pi+8 \pi-2 \pi}{12}=\frac{9 \pi}{12}=\frac{3 \pi}{4} \)
12. Find the values of the following:
\( \operatorname{Cos}^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right) \)
\( \operatorname{Cos}^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right) \)
Answer
Let \( \cos ^{-1}\left(\frac{1}{2}\right)=x \)
Then, we get,
\( \operatorname{Cos} x=\frac{1}{2}=\cos \frac{\pi}{3} \)
We know that range of the principle value branch of \( \cos ^{-1} \) is \( [0, \pi] \)
Therefore \( \cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3} \)
Let \( \sin ^{-1}\left(\frac{1}{2}\right)=y \) then \( \sin y=\frac{1}{2}=\sin \left(\frac{\pi}{6}\right) \)
We know that range of the principle value branch of \( \sin ^{-1} \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)
Therefore \( \sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6} \)
Now,
\( \operatorname{Cos}^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}+2 \times \frac{\pi}{6}=\frac{2 \pi}{3} \)
13. Find the values of the following:
\( \sin ^{-1} x=y \), then
A. \( 0 \leq y \leq \pi \)
B. \( -\frac{\pi}{2} \leq y \leq \frac{\pi}{2} \)
C. \( 0 < y < \pi \)
D. \( -\frac{\pi}{2} < y < \frac{\pi}{2} \)
\( \sin ^{-1} x=y \), then
A. \( 0 \leq y \leq \pi \)
B. \( -\frac{\pi}{2} \leq y \leq \frac{\pi}{2} \)
C. \( 0 < y < \pi \)
D. \( -\frac{\pi}{2} < y < \frac{\pi}{2} \)
Answer
\(
\sin ^{-1} x=y
\)
We know that range of the principle value branch of \( \sin ^{-1} \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)
Therefore, \( -\frac{\pi}{2} \leq y \leq \frac{\pi}{2} \)
Hence, the option (B) is correct.
14. Find the values of the following:
\( \tan ^{-1} \sqrt{3}-\sec ^{-1}(-2) \) is equal to
A. \( \pi \)
B. \( -\frac{\pi}{3} \)
C. \( \frac{\pi}{3} \)
D. \( \frac{2 \pi}{3} \)
\( \tan ^{-1} \sqrt{3}-\sec ^{-1}(-2) \) is equal to
A. \( \pi \)
B. \( -\frac{\pi}{3} \)
C. \( \frac{\pi}{3} \)
D. \( \frac{2 \pi}{3} \)
Answer
Let us take
\( \tan ^{-1}(\sqrt{3})=x \) Then we get,
\( \tan x=\sqrt{3}=\tan \frac{\pi}{3} \)
We know that range of the principle value branch of \( \tan ^{-1} \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)
Therefore, \( \tan ^{-1}(\sqrt{3})=\frac{\pi}{3} \)
Let \( \sec ^{-1}(-2)=y \) Then we get,
\( \operatorname{Sec} y=-2=-\sec \frac{\pi}{3}=\sec \left(\pi-\frac{\pi}{3}\right)=\sec \left(\frac{2 \pi}{3}\right) \)
We know that range of the principle value branch of \( \sec ^{-1} \) is \( [0, \pi]-\left\{\frac{\pi}{2}\right\} \)
Therefore, \( \sec ^{-1}(-2)=\frac{2 \pi}{3} \)
Now,
\(
\tan ^{-1} \sqrt{3}-\sec ^{-1}(-2)=\frac{\pi}{3}-\frac{2 \pi}{3}=-\frac{\pi}{3}
\)
Hence, the option (B) is correct.
