Chapter 2: Inverse Trigonometric Functions

Ans: To Prove $3{\sin ^{ - 1}}x = {\sin ^{ - 1}}\left( {3x - 4{x^3}} \right)$, where $x \in \left[ { - \dfrac{1}{2},\dfrac{1}{2}} \right]$

Let $x = \sin \theta .$ Then, ${\sin ^{ - 1}}x = \theta $.

We have given that,

R.H.S ${\sin ^{ - 1}}\left( {3x - 4{x^3}} \right) = {\sin ^{ - 1}}\left( {3\sin \theta  - 4{{\sin }^3}\theta } \right)$

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

$ = 3\theta $

$ = 3{\sin ^{ - 1}}x = L.H.S$

Hence proved.

Ans: To Prove $3{\cos ^{ - 1}}x = {\cos ^{ - 1}}\left( {4{x^3} - 3x} \right),x \in \left[ {\dfrac{1}{2},1} \right]$

Let $x = \cos \theta $. Then, ${\cos ^{ - 1}}x = \theta $

We have given that,

R.H S =${\cos ^{ - 1}}\left( {4{x^3} - 3x} \right)$

$ = {\cos ^{ - 1}}\left( {4{{\cos }^3}\theta  - 3\cos \theta } \right)$

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

$ = 3\theta $

$ = 3{\cos ^{ - 1}}x = L.H.S$

Hence proved.

Ans: ${\tan ^{ - 1}}\dfrac{{\sqrt {1 + {x^2}}  - 1}}{x}$

By putting $x = \tan \theta  \Rightarrow \theta  = {\tan ^{ - 1}}x$

$\therefore {\tan ^{ - 1}}\dfrac{{\sqrt {1 + {x^2}}  - 1}}{x} = {\tan ^{ - 1}}\dfrac{{\sqrt {1 + {{\tan }^2}\theta }  - 1}}{{\tan \theta }}$

$ = {\tan ^{ - 1}}\left( {\dfrac{{\sec \theta  - 1}}{{\tan \theta }}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{1 - \cos \theta }}{{\sin \theta }}} \right)$

${\tan ^{ - 1}}\left( {\dfrac{{2{{\sin }^2}\left( {\dfrac{\theta }{2}} \right)}}{{2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}}} \right)$

$ = {\tan ^{ - 1}}\left( {\tan \dfrac{\theta }{2}} \right) = \dfrac{\theta }{2} = \dfrac{1}{2}{\tan ^{ - 1}}x$

Ans: Given that,${\tan ^{ - 1}}\left( {\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right),x < \pi $

${\tan ^{ - 1}}\left( {\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right)$

$ = {\tan ^{ - 1}}\left( {\sqrt {\dfrac{{2{{\sin }^2}\dfrac{x}{2}}}{{2{{\cos }^2}\dfrac{x}{2}}}} } \right)$

$ = {\tan ^{ - 1}}\left( {\dfrac{{\sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}}} \right)$

$ = {\tan ^{ - 1}}\left( {\tan \dfrac{x}{2}} \right)$

$ = \dfrac{x}{2}$

Ans: We have given that, ${\tan ^{ - 1}}\left( {\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}}} \right)$

$ = {\tan ^{ - 1}}\left( {\dfrac{{1 - \left( {\dfrac{{\sin x}}{{\cos x}}} \right)}}{{1 + \left( {\dfrac{{\sin x}}{{\cos x}}} \right)}}} \right)$

$ = {\tan ^{ - 1}}\left( {\dfrac{{1 - \tan x}}{{1 + \tan x}}} \right)$

$ = {\tan ^{ - 1}}(1) - {\tan ^{ - 1}}(\tan x)\quad \left[ {\because \quad \dfrac{{ - y}}{{x - xy}} = {{\tan }^{ - 1}}x - {{\tan }^{ - 1}}y} \right]$

$ = \dfrac{\pi }{4} - x$

Ans: We have given that,${\tan ^{ - 1}}\dfrac{x}{{\sqrt {{a^2} - {x^2}} }}$

Let’s consider, $x = a\sin \theta  \Rightarrow \dfrac{x}{a} = \sin \theta  \Rightarrow {\sin ^{ - 1}}\left( {\dfrac{x}{a}} \right)$

$\therefore {\tan ^{ - 1}}\dfrac{x}{{\sqrt {{a^2} - {x^2}} }}$

$ = {\tan ^{ - 1}}\left( {\dfrac{{a\sin \theta }}{{\sqrt {{a^2} - {a^2}{{\sin }^2}\theta } }}} \right)$

$ = {\tan ^{ - 1}}\left( {\dfrac{{a\sin \theta }}{{a\sqrt {1 - {{\sin }^2}\theta } }}} \right)$

$ = {\tan ^{ - 1}}\left( {\dfrac{{a\sin \theta }}{{a\cos \theta }}} \right)$

$ = {\tan ^{ - 1}}(\tan \theta ) = \theta  = {\sin ^{ - 1}}\dfrac{x}{a}$

Ans: Consider, ${\tan ^{ - 1}}\left( {\dfrac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right)$

Let’s consider, $x = a\tan \theta  \Rightarrow \dfrac{x}{a} = \tan \theta  \Rightarrow \theta  = {\tan ^{ - 1}}\left( {\dfrac{x}{a}} \right)$

${\tan ^{ - 1}}\left( {\dfrac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right)$

$ = {\tan ^{ - 1}}\left( {\dfrac{{3{a^2} \cdot a\tan \theta  - {a^3}{{\tan }^3}\theta }}{{{a^3} - 3a \cdot {a^2}{{\tan }^2}\theta }}} \right)$

$ = {\tan ^{ - 1}}\left( {\dfrac{{3{a^3}\tan \theta  - {a^3}{{\tan }^3}\theta }}{{{a^3} - 3{a^3}{{\tan }^2}\theta }}} \right)$

$ = {\tan ^{ - 1}}(\tan 3\theta )$

$ = 3\theta $

$ = 3{\tan ^{ - 1}}\dfrac{x}{a}$

Ans: Let’s consider, ${\sin ^{ - 1}}\dfrac{1}{2} = x$

Then $\sin x = \dfrac{1}{2} = \sin \left( {\dfrac{\pi }{6}} \right)$

$\therefore {\sin ^{ - 1}}\dfrac{1}{2} = \dfrac{\pi }{6}$

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

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

$ = {\tan ^{ - 1}}\left[ {2\cos \dfrac{\pi }{3}} \right]$

$ = {\tan ^{ - 1}}\left[ {2 \times \dfrac{1}{2}} \right]$

$ = {\tan ^{ - 1}}1 = \dfrac{\pi }{4}$

Ans: Let consider, $x = \tan \theta $.

Then, $\theta  = {\tan ^{ - 1}}x$.

$\therefore {\sin ^{ - 1}}\dfrac{{2x}}{{1 + {x^2}}} = {\sin ^{ - 1}}\left( {\dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right)$

$ = {\sin ^{ - 1}}(\sin 2\theta )$

$ = 2\theta $

$ = 2{\tan ^{ - 1}}x$

Let’s assume, $y = \tan \theta .$ Then, $\theta  = {\tan ^{ - 1}}y$.

$\therefore {\cos ^{ - 1}}\left( {\dfrac{{1 - {y^2}}}{{1 + {y^2}}}} \right)$

$ = {\cos ^{ - 1}}\left( {\dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right)$

$ = {\cos ^{ - 1}}(\cos 2\theta )$

$ = 2\theta  = 2{\tan ^{ - 1}}y$

$\therefore \tan \dfrac{1}{2}\left[ {{{\sin }^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) + {{\cos }^{ - 1}}\left( {\dfrac{{1 - {y^2}}}{{1 + {y^2}}}} \right)} \right]$

$ = \tan \dfrac{1}{2}\left[ {2{{\tan }^{ - 1}}x + 2{{\tan }^{ - 1}}y} \right]$

$\left[ {As,{{\tan }^{ - 1}}x + {{\tan }^{ - 1}}y = {{\tan }^{ - 1}}\dfrac{{x + y}}{{1 - xy}}} \right]$

$ = \tan \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)} \right]$

$ = \dfrac{{x + y}}{{1 - xy}}$

Ans: Let’s consider, ${\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}{3}} \right)$

As we know that ${\sin ^{ - 1}}(\sin x) = x$

If $x \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$, which is the principal value branch of ${\sin ^{ - 1}}x$.

Here, $\dfrac{{2\pi }}{3} \notin \left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]$

Now, ${\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}{3}} \right)$ can be written as:

${\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}{3}} \right)$

$ = {\sin ^{ - 1}}\left[ {\sin \left( {\pi  - \dfrac{{2\pi }}{3}} \right)} \right]$

$ = {\sin ^{ - 1}}\left( {\sin \dfrac{\pi }{3}} \right)$, where $\dfrac{\pi }{3} \in \left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]$

$\therefore {\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}{3}} \right) = {\sin ^{ - 1}}\left[ {\sin \dfrac{\pi }{3}} \right] = \dfrac{\pi }{3}$