Class 12 Maths Chapter 2 Inverse Trigonometric Functions – Exercise 2.2 NCERT Solutions

Introduction

Exercise 2.2 focuses on properties and simplifications of inverse trigonometric functions. Students learn how to apply identities, prove results, and evaluate expressions involving multiple inverse trigonometric functions. This exercise is essential for mastering integration techniques and advanced trigonometry.

Formulas Used

1. Basic Identities:

$${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2},{\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2}$$

2. Conversion Identities:

$${\tan }^{-1}\left(\frac{1}{x}\right)={\cot }^{-1}x,x>0$$

3. Addition Formula:

$${\tan }^{-1}a+{\tan }^{-1}b={\tan }^{-1}\left(\frac{a+b}{1-ab}\right),ab<1$$

4. Subtraction Formula:

$${\tan }^{-1}a-{\tan }^{-1}b={\tan }^{-1}\left(\frac{a-b}{1+ab}\right)$$

Students Frequently Make Mistakes

• Forgetting principal value ranges.

• Misapplying addition formula without checking condition $ab<1$.

• Confusing reciprocal identities.

• Ignoring quadrant rules in simplification.

• Errors in proving identities step by step.

NCERT Questions with Step‑by‑Step Solutions (10)

Q1. Prove that ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$. Let $\theta ={\sin }^{-1}x$. Then $\sin \theta =x$. So, ${\cos }^{-1}x=\frac{\pi }{2}-\theta$. Hence, ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$. ✔

Q2. Prove that ${\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2}$. Let $\theta ={\tan }^{-1}x$. Then $\tan \theta =x$. So, ${\cot }^{-1}x=\frac{\pi }{2}-\theta$. Hence, sum = $\frac{\pi }{2}$. ✔

Q3. Evaluate ${\tan }^{-1}1+{\tan }^{-1}2$. Using formula:

$${\tan }^{-1}1+{\tan }^{-1}2={\tan }^{-1}\left(\frac{1+2}{1-1\cdot 2}\right)={\tan }^{-1}\left(\frac{3}{-1}\right)={\tan }^{-1}(-3)$$

Principal value: $-{\tan }^{-1}3$.

Q4. Evaluate ${\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{3}$.

$$={\tan }^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\cdot \frac{1}{3}}\right)={\tan }^{-1}\left(\frac{\frac{5}{6}}{1-\frac{1}{6}}\right)={\tan }^{-1}(\frac{5}{6}\cdot \frac{6}{5})={\tan }^{-1}(1)=\frac{\pi }{4}$$

Q5. Evaluate ${\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\frac{5}{12}$.

$$={\tan }^{-1}\left(\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}\cdot \frac{5}{12}}\right)={\tan }^{-1}\left(\frac{\frac{9+5}{12}}{1-\frac{15}{48}}\right)={\tan }^{-1}(\frac{14}{12}\cdot \frac{48}{33})={\tan }^{-1}\left(\frac{56}{33}\right)$$

Q6. Prove that ${\tan }^{-1}a+{\tan }^{-1}b={\tan }^{-1}\left(\frac{a+b}{1-ab}\right)$. Let $\theta ={\tan }^{-1}a,\varphi ={\tan }^{-1}b$. Then $\tan \theta =a,\tan \varphi =b$. So, $\tan (\theta +\varphi )=\frac{a+b}{1-ab}$. Hence, $\theta +\varphi ={\tan }^{-1}\left(\frac{a+b}{1-ab}\right)$. ✔

Q7. Prove that ${\tan }^{-1}a-{\tan }^{-1}b={\tan }^{-1}\left(\frac{a-b}{1+ab}\right)$. Similar proof using subtraction formula. ✔

Q8. Evaluate ${\tan }^{-1}2-{\tan }^{-1}1$.

$$={\tan }^{-1}\left(\frac{2-1}{1+2\cdot 1}\right)={\tan }^{-1}\left(\frac{1}{3}\right)$$

Q9. Evaluate ${\tan }^{-1}\frac{1}{\sqrt{3}}+{\tan }^{-1}\sqrt{3}$.

$$={\tan }^{-1}\left(\frac{\frac{1}{\sqrt{3}}+\sqrt{3}}{1-\frac{1}{\sqrt{3}}\cdot \sqrt{3}}\right)={\tan }^{-1}\left(\frac{\frac{1+3}{\sqrt{3}}}{0}\right)={\tan }^{-1}(\mathrm{\infty })=\frac{\pi }{2}$$

Q10. Evaluate ${\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{2}{3}$.

$$={\tan }^{-1}\left(\frac{\frac{1}{2}+\frac{2}{3}}{1-\frac{1}{2}\cdot \frac{2}{3}}\right)={\tan }^{-1}\left(\frac{\frac{7}{6}}{1-\frac{1}{3}}\right)={\tan }^{-1}(\frac{7}{6}\cdot \frac{3}{2})={\tan }^{-1}\left(\frac{7}{4}\right)$$

FAQs (10)

FAQ1. What is principal value of ${\sin }^{-1}x$? $[-\frac{\pi }{2},\frac{\pi }{2}]$.

FAQ2. What is principal value of ${\cos }^{-1}x$? $[0,\pi ]$.

FAQ3. What is principal value of ${\tan }^{-1}x$? $(-\frac{\pi }{2},\frac{\pi }{2})$.

FAQ4. What is identity of ${\sin }^{-1}x+{\cos }^{-1}x$? $\frac{\pi }{2}$.

FAQ5. What is identity of ${\tan }^{-1}x+{\cot }^{-1}x$? $\frac{\pi }{2}$.

FAQ6. What is formula for ${\tan }^{-1}a+{\tan }^{-1}b$? ${\tan }^{-1}\left(\frac{a+b}{1-ab}\right)$.

FAQ7. What is formula for ${\tan }^{-1}a-{\tan }^{-1}b$? ${\tan }^{-1}\left(\frac{a-b}{1+ab}\right)$.

FAQ8. Why check condition $ab<1$? To ensure sum lies in principal value range.

FAQ9. Is ${\sin }^{-1}x=\frac{1}{\sin x}$? No, it is inverse function, not reciprocal.

FAQ10. Why is Exercise 2.2 important? It builds foundation for integration involving inverse trigonometric functions.

Conclusion

Exercise 2.2 has 10 solved questions and 10 FAQs that strengthen your understanding of identities and simplifications of inverse trigonometric functions. This builds the foundation for calculus in Class 12 Maths.

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