# Chapter 2: Inverse Trigonometric Functions

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

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### 1 Comment

1. Harshit raj
March 9, 2018

Nice we get full help.thanks byju’s community