Question

Solve the equation
$\cos \left({\tan }^{-1}x\right)=\sin \left({\cot }^{-1}\frac{3}{4}\right)$

Answer 1

Given : $\cos \left({\tan }^{-1}x\right)=\sin \left({\cot }^{-1}\frac{3}{4}\right)$
Taking $LHS$,
$LHS=\cos \left({\tan }^{-1}x\right)$
$LHS=\cos \left({\cos }^{-1}\frac{1}{\sqrt{x^2+1}}\right)$
$[\because {\tan }^{-1}x={\cos }^{-1}\frac{1}{\sqrt{x^2+1}}]$
$LHS=\frac{1}{\sqrt{x^2+1}}$
$[\because \cos ({\cos }^{-1}x)=x:xϵ[-1,1\left]\right]$
Taking $RHS$,
$RHS=\sin \left({\cot }^{-1}\frac{3}{4}\right)$
$RHS=\sin \left({\sin }^{-1}\frac{4}{5}\right)$
$[\because {\cot }^{-1}\frac{x}{y}={\sin }^{-1}\frac{y}{\sqrt{x^2+y^2}}]$
$RHS=\frac{4}{5}$
$[\because \sin ({\sin }^{-1}x)=x:xϵ[-1,1\left]\right]$
As $LHS=RHS$,
$\Rightarrow \frac{1}{\sqrt{x^2+1}}=\frac{4}{5}$
$\Rightarrow 16+16x^2=25$ [Squaring both sides]
$\Rightarrow x^2=\frac{9}{16}$
$\Rightarrow x=\pm \frac{3}{4}$