Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

Given,

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

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

So, $\cos \left({\sin }^{-1}\left(x\cot \left({\tan }^{-1}\left(\cos \left({\sin }^{-1}x\right)\right)\right)\right)\right)=k$

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

And now solving  $\cot \left({\tan }^{-1}\sqrt{1-x^2}\right)={\mathrm{cotcot}}^{-1}\left(\sqrt{\frac{1}{\sqrt{1-x^2}}}\right)=\frac{1}{\sqrt{1-x^2}}$

So, $\cos \left({\sin }^{-1}\left(x\cot \left({\tan }^{-1}\sqrt{1-x^2}\right)\right)\right)=k$ becomes 

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

Now solving $\cos \left({\sin }^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\right)=\frac{\sqrt{1-2x^2}}{\sqrt{1-x^2}}$

So, $\frac{\sqrt{1-2x^2}}{\sqrt{1-x^2}}=k$

$\Rightarrow 1-2x^2=k^2\left(1-x^2\right)$

$\Rightarrow \left(k^2-2\right)x^2=k^2-1$

$\Rightarrow x^2=\frac{k^2-1}{k^2-2}$

So, roots are $\alpha =\sqrt{\frac{k^2-1}{k^2-2}}\Rightarrow {\alpha }^2=\frac{k^2-1}{k^2-2}$

And $\beta =-\sqrt{\frac{k^2-1}{k^2-2}}\Rightarrow {\beta }^2=\frac{k^2-1}{k^2-2}$

Now finding $\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}=2\left(\frac{k^2-2}{k^2-1}\right)$ and $\frac{\alpha }{\beta }=-1$

So, sum of roots of $x^2-bx-5=0$ will be $=\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{\alpha }{\beta }=b$

$\Rightarrow \frac{2\left(k^2-2\right)}{k^2-1}-1=b \cdots \left(1\right)$

Product of roots of $x^2-bx-5=0$ will be $=\left(\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}\right)\frac{\alpha }{\beta }=-5$

$\Rightarrow \frac{2\left(k^2-2\right)}{k^2-1}\left(-1\right)=-5$

$\Rightarrow 2k^2-4=5k^2-5$

$\Rightarrow 3k^2=1\Rightarrow k^2=\frac{1}{3} \cdots \mathrm{Put} \mathrm{in} \left(1\right)$

$\Rightarrow b=\frac{2\left(k^2-2\right)}{k^2-1}-1=5-1=4$

$\frac{b}{k^2}=\frac{4}{\frac{1}{3}}=12$