Question

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

Answer 1

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

We know that,${\cos }^{-1}x={\tan }^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)$ and ${\tan }^{-1}x={\sin }^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)$

Now,

$\tan \left({\cos }^{-1}\frac{1}{x}\right)=\sin \left(ta{n}^{-1}x\right)$

$\Rightarrow \tan {\tan }^{-1}\left(\frac{\sqrt{1-\frac{1}{x^2}}}{\frac{1}{x}}\right)=\sin {\sin }^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)$

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

$\Rightarrow \sqrt{x^2-1^2}.\sqrt{x^2+1^2}=x$

$\Rightarrow \sqrt{x^4-1^4}=x$

Squaring both side and we get

$\Rightarrow x^4-1=x^2$

$\Rightarrow x^4-1-x^2=0$

$\Rightarrow x^4-x^2-1=0$

Let $x^2=y$

$\Rightarrow y^2-y-1=0$

Comparing that, $a{y}^2+by+c=0$

$a=1,b=-1,c=-1$

Using quadratic equation formula, and we get

$y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$y=\frac{-(-1)\pm \sqrt{{(-1)}^2-4\times 1\times (-1)}}{2\times 1}$

$y=\frac{1\pm \sqrt{5}}{2}$

And

$y=x^2=\frac{1\pm \sqrt{5}}{2}$

$x=\sqrt{\frac{1\pm \sqrt{5}}{2}}$

Hence, this is the answer.