NCERT Solutions For Class 12 Maths Chapter 2

NCERT Solutions Class 12 Maths Inverse Trigonometric Functions

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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