Inverse Trigonometric Functions

Given the function $ y = \tan^{-1}\left(\frac{\sqrt{x}-x}{1+x^{\frac{3}{2}}}\right) $.

We can rewrite this expression using an identity for inverse tangent:

$ y = \tan^{-1}\left(\frac{\sqrt{x}-x}{1+\sqrt{x} \cdot x}\right) $

This can be expressed as:

$ y = \tan^{-1}(\sqrt{x}) - \tan^{-1}(x) $

This transformation uses the identity:

$ \tan^{-1}\left(\frac{a-b}{1+ab}\right) = \tan^{-1}(a) - \tan^{-1}(b) $

To find the derivative $ y' $ with respect to $ x $, we use the differentiation of inverse tangent functions:

$ y' = \frac{d}{dx}[\tan^{-1}(\sqrt{x}) - \tan^{-1}(x)] $

The derivative is calculated as follows:

$ y' = \frac{1}{1+(\sqrt{x})^2} \cdot \frac{1}{2\sqrt{x}} - \frac{1}{1+x^2} $

Now, evaluate $ y'(1) $:

$ y'(1) = \frac{1}{1+1} \cdot \frac{1}{2\cdot1} - \frac{1}{1+1} = \frac{1}{2} \cdot \frac{1}{2} - \frac{1}{2} = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} $

Thus, the value of $ y'(1) $ is $ -\frac{1}{4} $.

Given, $\cot ^{-1}(\sqrt{\cos \alpha})-\tan ^{-1} \sqrt{\cos \alpha}=x$ ... (i)

We know that $\cot ^{-1}(\sqrt{\cos } \alpha)+\tan ^{-1} \sqrt{\cos \alpha}=\frac{\pi}{2}$ .... (ii)

On adding Eqs. (i) and (ii), we get

$\begin{aligned} & 2 \cot ^{-1} \sqrt{\cos \alpha}=\frac{\pi}{2}+x \\ & \Rightarrow \sqrt{\cos } \alpha=\cot \left(\frac{\pi}{4}+\frac{x}{2}\right)=\frac{\cot \frac{x}{2}-1}{1+\cot \frac{x}{2}} \end{aligned}$

$\Rightarrow \sqrt{\cos \alpha}=\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}$

On squaring both sides, we get

$\cos \alpha=\frac{\cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}-2 \sin \frac{x}{2} \cos \frac{x}{2}}{\cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}}$

$\begin{aligned} & \Rightarrow \cos \alpha=\frac{1-\sin x}{1+\sin x} \\ & \Rightarrow \frac{1-\tan ^2 \frac{\alpha}{2}}{1+\tan ^2 \frac{\alpha}{2}}=\frac{1-\sin x}{1+\sin x} \end{aligned}$

On applying componendo and dividendo rule, we get

$\begin{aligned} & \frac{1-\tan ^2 \frac{\alpha}{2}-1-\tan ^2 \frac{\alpha}{2}}{1-\tan ^2 \frac{\alpha}{2}+1+\tan ^2 \frac{\alpha}{2}}=\frac{1-\sin x-1-\sin x}{1-\sin x+1+\sin x} \\ \Rightarrow \quad & \tan ^2 \frac{\alpha}{2}=\sin x \\ \Rightarrow \quad & \sin x=\tan ^2 \frac{\alpha}{2} \end{aligned}$

$y = {\cos^{ - 1}}\left( {{7 \over 2}(1 + \cos 2x) + \sqrt {({{\sin }^2}x - 48{{\cos }^2}x)} \sin x} \right)$

$ = {\cos ^{ - 1}}((7\cos x)(\cos x) + \sqrt {1 - 49{{\cos }^2}x} \sqrt {1 - {{\cos }^2}x} )$

$ = {\cos ^{ - 1}}(\cos x) - {\cos ^{ - 1}}(7\cos x)$ [$\because$ $\cos x < 7\cos x$]

$ = x - {\cos ^{ - 1}}(7\cos x)$

Let $I = {({\sin ^{ - 1}}x)^3} + {({\cos ^{ - 1}}x)^3}$

$ = ({\sin ^{ - 1}}x + {\cos ^{ - 1}}x)[{({\sin ^{ - 1}}x)^2} + {({\cos ^{ - 1}}x)^2} - ({\sin ^{ - 1}}x)({\cos ^{ - 1}}x)]$

$ = {\pi \over 2}\left[ {{{({{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x)}^2} - 3{{\sin }^{ - 1}}x\left( {{\pi \over 2} - {{\sin }^{ - 1}}x} \right)} \right]$

$ = {\pi \over 2}\left( {{{{\pi ^2}} \over 4} - {{3\pi } \over 2}{{\sin }^{ - 1}}x + 3{{({{\sin }^{ - 1}}x)}^2}} \right)$

$ = {\pi \over 2}\left[ {3\left( {{{({{\sin }^{ - 1}}x)}^2} - {\pi \over 2}({{\sin }^1}x) + {{{\pi ^2}} \over {16}} - {{{\pi ^2}} \over {16}}} \right) + {{{\pi ^2}} \over 4}} \right]$

$ = {\pi \over 2}\left( {{{{\pi ^2}} \over {16}} + 3{{\left( {{{\sin }^{ - 1}}x - {\pi \over 4}} \right)}^2}} \right)$

Now, ${\pi \over 4} \ge {\sin ^{ - 1}}x - {\pi \over 4} \ge - {{3\pi } \over 4}$

$ \Rightarrow {{9{\pi ^2}} \over {16}} \ge {\left( {{{\sin }^{ - 1}}x - {\pi \over 4}} \right)^2} \ge 0 \Rightarrow I \ge {\pi \over 2}\,.\,{{{\pi ^2}} \over {16}} = {{{\pi ^3}} \over {32}}$