# NCERT Solutions For Class 12 Maths Chapter 2

## NCERT Solutions Class 12 Maths Inverse Trigonometric Functions

BYJU’S, India’s largest learning app provide free NCERT solutions for all the classes from class 6 to 12. NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions, available here are free in PDF format.

All the solutions provided in these materials are prepared by the subject experts and is framed as per the latest and updated CBSE syllabus. All solved questions available in NCERT Solutions for class 12 maths chapter 2 include detailed solutions to the questions of the NCERT textbooks including both objective and short and long type questions.

NCERT Solutions for class 12 maths chapter 2 pdf help students to have a quick revision of a complete chapter along with its important questions.

### NCERT Solutions Class 12 Maths Chapter 2 Exercises

Exercise 2.1

Q1. Find principal value for $$\sin^{-1}\left ( -\frac{1}{2} \right )$$

Soln:

Let $$\sin^{-1}\left ( -\frac{1}{2} \right )$$ = a, then

$$\sin^{a} = -\frac{1}{2} = – \sin\frac{ \pi}{6} = \sin ( – \frac{ \pi}{6} )$$

We know,

The principal value branch range for sin-1 is $$\left [ -\frac{ \pi}{2}, \frac{ \pi}{2} \right ]$$ and $$\sin ( -\frac{ \pi}{6} ) = – \frac{1}{2}$$

Therefore principal value for $$\sin^{-1}\left ( -\frac{1}{2} \right ) \; is \; – \frac{ \pi}{6}$$

Q2. Find principal value for $$\cos^{-1}\left ( – \frac{\sqrt{3}}{2} \right )$$

Soln:

Let $$\cos^{-1}\left ( – \frac{\sqrt{3}}{2} \right )$$ = a, then

$$\cos a = \frac{\sqrt{3}}{2} = \cos (\frac{\pi}{6})$$

We know,

The principal value branch range for cos-1 is $$\left [ 0 , \pi \right ]$$ and $$\cos (\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$$

Therefore, principal value for $$\cos^{-1}\left ( – \frac{\sqrt{3}}{2} \right ) \; is \; \frac{\pi}{6}$$

Q3. Find principal value for cosec-1 (2)

Soln:

Let cosec-1 (2) = a. Then, cosec a = 2 = cosec $$(\frac{\pi}{6} )$$

We know,

The principal value branch range for cosec-1 is $$\left [ -\frac{\pi}{2}, \frac{\pi}{2}\right ] – {0}$$ and cosec$$(\frac{\pi}{6} )$$ = 2

Therefore, principal value for cosec‑1 (2) is $$\frac{\pi}{6}$$

Q4. Find principal value for $$\tan^{-1} \left ( – \sqrt{3} \right )$$

Soln:

Let $$\tan^{-1} \left ( – \sqrt{3} \right ) = a$$

Then, $$\tan = – \sqrt{3} = – \tan \frac{\pi}{3} \tan (- \frac{\pi}{3})$$

We know,

The principal value branch range for $$\tan^{-1} \; is \; \left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ] \; and \; \tan\left ( -\frac{\pi}{3} \right ) = -\sqrt{3}$$

Therefore, principal value for $$\tan^{-1} \left ( – \sqrt{3} \right ) \; is \; -\frac{\pi}{3}$$

Q5. Find principal value for $$\cos^{-1}\left ( -\frac{1}{2} \right )$$

Soln:

Let $$\cos^{-1}\left ( -\frac{1}{2} \right )$$ = a,

Then $$\cos a = -\frac{1}{2} = -cos \frac{\pi}{3} = \cos ( \pi – \frac{\pi}{3} ) = \cos( \frac{2 \pi}{3} )$$

We know,

The principal value branch range for $$\cos ^{-1} \; is \; \left [ 0 , \pi \right ] \; and \; \cos \left ( \frac{2 \pi}{3} \right ) = – \frac{1}{2}$$

Therefore, principal value for $$\cos^{-1}\left ( -\frac{1}{2} \right ) \; is \; \frac{2 \pi}{3}$$

Q6. Find principal value for $$\tan^{-1} (-1)$$

Soln:

Let $$\tan^{-1} (-1) = a$$,

Then, tan a = -1 = $$-\tan ( \frac{\pi}{4} ) = \tan ( – \frac{\pi}{4} )$$

We know,

The principal value branch range for $$\tan^{-1} \; is \; \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right ) \; and \; \tan ( – \frac{\pi}{4} ) = -1$$

Therefore, principal value for $$\tan^{-1} (-1) \; is \; -\frac{\pi}{4}$$

Q7. Find principal value for $$\sec^{-1} \left ( \frac{2}{\sqrt{3}} \right )$$

Soln:

Let $$\sec^{-1} \left ( \frac{2}{\sqrt{3}} \right ) = a$$,

Then $$\sec a = \frac{2}{\sqrt{3}} = \sec (\frac{\pi}{6})$$

We know,

The principal value branch range for $$\sec^{-1} \; is \; \left [ 0 , \pi \right ] – \left \{ \frac{ \pi }{2} \right \} \; and \; \sec (\frac{\pi}{6}) = \frac{2}{\sqrt{3}}$$

Therefore, principal value for $$\sec^{-1} \left ( \frac{2}{\sqrt{3}} \right ) \; is ; \frac{\pi}{6}$$

Q8. Find principal value for $$\cot^{-1} \sqrt{3}$$

Soln:

Let $$\cot^{-1} \sqrt{3} = a$$,

Then $$\cot a = \sqrt {3} = \cot \left ( \frac{\pi}{6} \right )$$

We know,

The principal value branch range for cot­-1 is $$( 0 , \pi )$$ and $$\cot \left ( \frac{\pi}{6} \right ) = \sqrt{3}$$

Therefore, principal value for $$\cot^{-1} \sqrt{3} = \frac{\pi}{6}$$

Q9. Find principal value for $$\cos^{-1} \left ( – \frac{1}{\sqrt{2}} \right )$$

Soln:

Let $$\cos^{-1} \left ( – \frac{1}{\sqrt{2}} \right ) = a$$

Then $$\cos a = \frac{-1}{\sqrt{2}} = – \cos \left ( \frac{\pi}{4} \right ) = \cos \left ( \pi – \frac{\pi}{4} \right ) = \cos \left ( \frac{3 \pi}{4} \right )$$

We know,

The principal value branch range for cos‑1 is $$[0 , \pi] \; and \; \cos \left ( \frac{3 \pi}{4} \right ) = -\frac{1}{\sqrt{2}}$$

Therefore, principal value for $$\cos^{-1} \left ( – \frac{1}{\sqrt{2}} \right ) \; is \; \frac{3 \pi }{4}$$

Q10. Find principal value for cosec-1 $$\left ( -\sqrt{2} \right )$$

Soln:

Let cosec-1$$\left ( -\sqrt{2} \right )$$ = a, Then

cosec a = $$-\sqrt{2}$$ = -cosec$$\left ( \frac{\pi}{4}\right )$$ = cosec $$\left ( -\frac{\pi}{4}\right )$$

We know,

The principal value branch range for cosec-1 is $$\left [ -\frac{\pi}{2} , \frac{\pi}{2} \right ] – \left \{ 0 \right \}$$ and cosec$$\frac{-\pi}{4} = -\sqrt{2}$$

Therefore, principal value for cosec-1 $$\left ( -\sqrt{2} \right ) \; is \; -\frac{\pi}{4}$$

Q11. Solve $$\tan ^{-1}(1) + \cos^{-1} \left ( -\frac{1}{2} \right ) + \sin ^{-1}\left ( -\frac{1}{2} \right )$$

Soln:

Let $$\tan ^{-1}(1) = a$$, then

$$\tan a = 1 = \tan \frac{\pi}{4}$$

We know,

The principal value branch range for $$\tan ^{-1} \; is \; \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$$

$$\tan ^{-1}(1) = \frac{\pi}{4}$$

Let $$\cos^{-1} \left ( -\frac{1}{2} \right ) = b$$, then

$$\cos b = -\frac{1}{2} = -\cos \frac {\pi}{3} = \cos\left ( \pi – \frac{\pi}{3} \right ) = \cos\left ( \frac{2\pi}{3} \right )$$

We know,

The principal value branch range for cos-1 is $$[0 , \pi]$$

$$\cos ^{-1} \left ( -\frac{1}{2} \right ) = \frac{2 \pi }{3}$$

Let $$\sin^{-1}\left ( -\frac{1}{2} \right ) = c$$, then

$$\sin c = – \frac{1}{2} = – \sin \frac{\pi}{6} = \sin \left ( -\frac{\pi}{6} \right )$$

We know,

The principal value branch range for $$\sin ^{-1} \; is \; \left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]$$

$$\sin^{-1} \left ( -\frac{1}{2} \right ) = – \frac{\pi}{6}$$

Now

$$\tan ^{-1}(1) + \cos^{-1} \left ( -\frac{1}{2} \right ) + \sin ^{-1}\left ( -\frac{1}{2} \right )$$ $$= \frac{\pi}{4} + \frac{2\pi}{3} – \frac{\pi}{6} = \frac{3\pi + 8\pi – 2\pi}{12} = \frac{9\pi}{12} = \frac{3\pi}{4}$$

Q12. Solve $$\cos ^{-1} \left ( \frac{1}{2} \right ) + 2 \sin ^{-1} \left ( \frac{1}{2} \right )$$

Soln:

Let $$\cos ^{-1} \left ( \frac{1}{2} \right ) = a$$, then $$\cos a = \frac{1}{2} = \cos \frac{\pi}{3}$$

We know,

The principal value branch range for cos-1 is $$\left [0 , \pi \right ]$$

$$\cos ^{-1} \left ( \frac{1}{2} \right ) = \frac{\pi}{3}$$

Let $$\sin ^{-1} \left (- \frac{1}{2} \right ) = b$$, then $$\sin b = \frac{1}{2} = \sin \frac{\pi}{6}$$

We know,

The principal value branch range for $$\sin ^{-1} \; is \; \left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]$$

$$\sin ^{-1} \left ( \frac{1}{2} \right ) = \frac{\pi}{6}$$

Now,

$$\cos ^{-1} \left ( \frac{1}{2} \right ) + 2 \sin ^{-1} \left ( \frac{1}{2} \right )$$ $$= \frac{\pi}{3} + 2 \times \frac{\pi}{6} = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3}$$

Q13. If sin-1 a = b, then

(i) $$0 \leq b \leq \pi$$

(ii) $$-\frac{\pi}{2} \leq b \leq \frac{\pi}{2}$$

(iii) $$0 < b < \pi$$

(iv) $$-\frac{\pi}{2} < b < \frac{\pi}{2}$$

Soln:

Given sin-1 a = b

We know,

The principal value branch range for $$\sin ^{-1} \; is \; \left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]$$

Therefore, $$-\frac{\pi}{2} \leq b \leq \frac{\pi}{2}$$

Q14. The value of $$\tan ^{-1} \sqrt{3} – \sec ^{-1}(-2)$$ is

(i) $$\pi$$

(ii) $$– \frac{\pi}{3}$$

(iii) $$\frac{\pi}{3}$$

(iv) $$\frac{2 \pi}{3}$$

Soln:

Let $$\tan ^{-1} \sqrt{3} = a$$, then

$$\tan a = \sqrt{3} = \tan \frac{\pi}{3}$$

We know

The principal value branch range for $$\tan ^{-1} \; is \; \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$$

$$\tan ^{-1}\sqrt{3} = \frac{\pi}{3}$$

Let sec-1(-2) = b, then

sec b = -2 = $$– \sec \frac{\pi}{3} = \sec \left ( \pi – \frac{\pi}{3} \right ) = \sec \left ( \frac{2 \pi}{3} \right )$$

We know

The principal value branch range for sec-1 is $$[0 , \pi] – \left \{ \frac{\pi}{2} \right \}$$

$$\sec ^{-1}(-2) = \frac{2 \pi}{3}$$

Now,

$$\tan ^{-1} \sqrt{3} – \sec ^{-1}(-2) = \frac{\pi}{3} – \frac{2 \pi}{3} = – \frac{\pi}{3}$$

Hence option (ii) is correct

Exercise 2.2

Q1. Show that $$3 \sin ^{-1} = \sin ^{-1}(3x – 4x^{3}) , \; x \in \left [ -\frac{1}{2}, \frac{1}{2} \right ]$$

Soln:

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

Let sin­-1x = Ɵ, then x = sin Ɵ

We get,

RHS = $$\sin ^{-1} (3x – 4x^{3 }) = \sin ^{-1} (3 \sin \Theta – 4 \sin^{3} \Theta )\\$$

= $$\\\sin ^{-1} (\sin 3 \Theta) = 3 \Theta = 3 \sin^{-1}x$$

= LHS

Q2. Show that $$3 \cos ^{-1} x = cos ^{-1}(4x^{3} – 3x), x \in \left [ \frac{1}{2}, 1 \right ]$$

Soln:

To show: $$3 \cos ^{-1} x = cos ^{-1}(4x^{3} – 3x), x \in \left [ \frac{1}{2}, 1 \right ]$$

Let cos-1 x = Ɵ, then x = cos Ɵ

We get,

RHS = $$\cos ^{-1} (4x^{3} – 3x) = cos^{-1}(4cos^{3} \Theta – 3cos \Theta )$$

= $$\\\cos ^{-1} (cos 3 \Theta ) = 3 \Theta = 3 cos^{-1} x$$

= LHS

Q3. Show that $$tan ^{-1} \frac{2}{11} + \tan ^{-1} \frac{7}{24} = \tan ^{-1} \frac{1}{2}$$

Soln:

To show: $$tan ^{-1} \frac{2}{11} + \tan ^{-1} \frac{7}{24} = \tan ^{-1} \frac{1}{2}$$

LHS = $$tan ^{-1} \frac{2}{11} + \tan ^{-1} \frac{7}{24}$$

$$= tan ^{-1} \left ( \frac{\frac{2}{11} + \frac{7}{24}}{1 – \frac{2}{11} \times \frac{7}{24}} \right ) = \tan^{-1} \left (\frac{\frac{48 + 77}{11 \times 24}}{\frac{11 \times 24 – 14}{11 \times 24}} \right )\\$$

$$\\= tan ^{-1} \frac{48 + 77}{264 – 14} = \tan^{-1} \frac{125}{251} = \tan^{-1} \frac{1}{2}$$ = RHS

Q4. Show that $$2 \tan^{-1} \frac{1}{2} + \tan^{-1}\frac{1}{7} = \tan^{-1}\frac{31}{17}$$

Soln:

To show: $$2 \tan^{-1} \frac{1}{2} + \tan^{-1}\frac{1}{7} = \tan^{-1}\frac{31}{17}$$

LHS = $$2 \tan^{-1} \frac{1}{2} + \tan^{-1}\frac{1}{7}$$

$$= \tan^{-1} \left [ \frac{2 \times \frac{1}{2}}{1 – \left ( \frac{1}{2} \right )^{2}} \right ] + \tan ^{-1} \frac{1}{7} = \tan ^{-1} \frac{1}{\left ( \frac{3}{4} \right )} + \tan^{-1} \frac{1}{7}\\$$ $$\\= \tan^{-1} \frac{4}{3} + \tan^{-1} \frac{1}{7} = \tan^{-1}\left ( \frac{\frac{4}{3} + \frac{1}{7}}{1 – \frac{4}{3} \times \frac{1}{7}} \right )\\$$ $$\\= \tan^{-1} \left ( \frac{\frac{28 + 3}{3 \times 7}}{\frac{3 \times 7 -4}{3 \times 7}} \right ) = \tan^{-1} \frac{28 + 3}{21 – 4} = tan^{-1} \frac{31}{17} = RHS$$

Q5. Find simplest form for $$\tan^{-1}\frac{\sqrt{1 + a^{2}} – 1}{a}, \; a \neq 0$$

Soln:

Given $$\tan^{-1}\frac{\sqrt{1 + a^{2}} – 1}{a}$$

Let a = tan Ɵ

$$= \tan^{-1}\frac{\sqrt{1 + a^{2}} – 1}{a}$$ = $$\tan^{-1} \frac{\sqrt{1 + \tan^{2}\Theta } – 1}{\tan \Theta } \\$$

$$\\ = \tan^{-1} \left ( \frac{ \sec \Theta – 1 }{\tan \Theta } \right ) = \tan^{-1} \left ( \frac{1 – \cos \Theta }{\sin \Theta } \right )\\$$ $$\\\tan^{-1} \left ( \frac{2\sin^{2}\frac{\Theta }{2}}{2\sin\frac{\Theta }{2}\cos\frac{\Theta }{2}}\right ) = \tan^{-1}\left ( \tan \frac{\Theta }{2} \right )\\$$ $$\\= \frac{\Theta }{2} = \frac{1}{2}\tan^{-1}a$$

Q6. Find the simplest form for $$\tan^{-1}\frac{1}{\sqrt{a^{2}-1}}$$, |a|> 1

Soln:

Given $$\tan^{-1}\frac{1}{\sqrt{a^{2}-1}}$$

Let a = csc Ɵ

$$\tan^{-1}\frac{1}{\sqrt{a^{2}-1}} = \tan^{-1}\frac{1}{\sqrt{\csc^{2}\Theta -1}}$$ $$=\tan^{-1}\frac{1}{ \cot \Theta } = \tan^{-1} \tan \Theta = \Theta = \csc ^{-1}a$$ $$= \frac{\pi}{2} – sec^{-1}a$$

Q7. Find simplest form for $$\tan^{-1} \left ( \sqrt{\frac{1 – \cos a}{1 + \cos a}} \right ), a < \pi,$$

Soln:

Given $$\tan^{-1} \left ( \sqrt{\frac{1 – \cos a}{1 + \cos a}} \right )$$

Now,

$$\tan^{-1} \left ( \sqrt{\frac{1 – \cos a}{1 + \cos a}} \right ) = \tan^{-1} \left ( \sqrt{\frac{2 \sin^{2}\frac{x}{2}}{2 \cos^{2}\frac{x}{2}}} \right ) \\$$ $$\\\tan^{-1} \left ( \sqrt{\tan^{2}\frac{x}{2}} \right ) = \tan^{-1}\left ( \tan \frac{x}{2} \right ) = \frac{x}{2}$$

Q8. Find simplest form for $$\tan^{-1} \left ( \frac{\cos a – \sin a}{\cos a + \sin a} \right ), 0 < a < \pi$$

Soln:

Given $$\tan^{-1} \left ( \frac{\cos a – \sin a}{\cos a + \sin a} \right )$$

Now,

$$\tan^{-1} \left ( \frac{\cos a – \sin a}{\cos a + \sin a} \right ) = \tan^{-1} \left ( \frac{1 – \frac{\sin a}{\cos a}}{1 + \frac{\sin a}{\cos a}} \right ) = \tan^{-1} \left ( \frac{1 – \tan a}{1 + \tan a} \right )\\$$

= $$\\\tan^{-1} \left ( \frac{1 – \tan a}{1 + 1.\tan a} \right ) = \tan^{-1} \left ( \frac{\tan \frac{\pi}{4} – \tan a}{1 + \tan \frac{\pi}{4}.\tan a}\right )\\$$

= $$\\\tan^{-1} \left [ \tan \left ( \frac{\pi}{4} – a \right )\right ] = \frac{\pi}{4} – a$$

Q9: Find simplest form for $$\tan^{-1} \frac{a}{\sqrt{x^{2} – a^{2}}}, \left | a \right | < x$$

Soln:

Given: $$\tan^{-1} \frac{a}{\sqrt{x^{2} – a^{2}}}$$

Let a = x sin Ɵ

$$\tan^{-1} \frac{a}{\sqrt{x^{2} – a^{2}}} = \tan^{-1} \left ( \frac{x\sin \Theta }{\sqrt{x^{2} – x^{2}\sin^{2}\Theta }} \right ) = \tan^{-1}\left ( \frac{x\sin \Theta }{x \sqrt{1 – \sin^{2}\Theta }} \right ) \\$$

= $$\\\tan^{-1} \left ( \frac{x \sin \Theta }{x \sin \Theta } \right ) = tan ^{-1} (\tan \Theta ) = \Theta = \sin ^{-1} \frac{a}{x}$$

Q10. Find simplest form for $$\tan^{-1} \left ( \frac{3x^{2}a – a^{3}}{x^{3} – 3xa^{2}} \right ) , x > 0; \frac{-x}{\sqrt{3}} \leq a\frac{x}{\sqrt{3}}$$

Soln:

Given $$\tan^{-1} \left ( \frac{3x^{2}a – a^{3}}{x^{3} – 3xa^{2}} \right )$$

Let a = x tan Ɵ

$$\tan^{-1} \left ( \frac{3x^{2}a – a^{3}}{x^{3} – 3xa^{2}} \right ) = \tan^{-1} \left ( \frac{3x^{2}.x \tan \Theta – x^{3}\tan^{3}\Theta }{x^{3} – 3x.x^{2}\tan^{2}\Theta } \right ) \\$$

=$$\\\tan^{-1} \left ( \frac{3x^{3} \tan \Theta – x^{3}\tan^{3}\Theta }{x^{3} – 3x^{3}\tan^{2}\Theta } \right ) = \tan^{-1} \left ( \frac{3 \tan \Theta – \tan^{3}\Theta }{1 – 3\tan^{2}\Theta } \right ) \\$$

= $$\tan^{-1} \left ( \tan 3 \Theta \right ) = 3 \Theta = 3 tan ^{-1} \frac{a}{x}$$

Q11. Solve $$\tan^{-1}\left [ 2\cos \left ( 2 \sin^{-1} \frac{1}{2} \right ) \right ]$$

Soln:

Given $$\tan^{-1}\left [ 2\cos \left ( 2 \sin^{-1} \frac{1}{2} \right ) \right ]$$

$$\tan^{-1}\left [ 2\cos \left ( 2 \sin^{-1} \frac{1}{2} \right ) \right ] = \tan^{-1}\left [ 2\cos \left ( 2 \sin^{-1} \left ( \sin \frac{\pi}{6} \right ) \right ) \right ] \\$$

= $$\\\tan^{-1}\left [ 2\cos \left ( 2 \times \frac{\pi}{6} \right ) \right ] = \tan^{-1} \left [ 2 \cos \left ( \frac{\pi}{3} \right ) \right ] = \tan^{-1} \left [ 2 \times \frac{1}{2} \right ]\\$$

= $$\tan^{-1}\left [ 1 \right ] = \frac{\pi}{4}$$

Q12. Solve $$\cot \left (\tan^{-1} x + \cot ^{-1} x \right )$$

Soln:

Given $$\cot \left (\tan^{-1} x + \cot ^{-1} x \right )$$

$$\cot \left (\tan^{-1} x + \cot ^{-1} x \right ) = \cot \left( \frac{\pi}{2} \right)$$

= 0