# NCERT Solutions For Class 12 Maths Chapter 2

## NCERT Solutions Class 12 Maths Inverse Trigonometric Functions

### Ncert Solutions For Class 12 Maths Chapter 2 PDF Free Download

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NCERT class 12 maths chapter 2 solutions 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

Class 12th is a very important phase of a students life. The marks scored in class 12 plays an important role while taking admission in colleges. The CBSE or Central Board of Secondary Education is responsible for conducting the board examination of class 12. To score good marks in the examination students should follow some good textbooks and study materials. NCERT textbooks are one of the best textbooks for the students of class 12. NCERT textbooks have all the topics covered which are necessary to study for class 2 board examination. Students are also advised to solve all the NCERT questions provided in the NCERT textbooks.

Mathematics is a subject which requires lots and lots of practice, so solving the previous year sample papers along with NCERT solutions will help the students to know about the types of questions asked in the examination as well as the marking scheme of each types of question.

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