Browse Questions

# If $\cot(\cos^{-1}x)=\sec\big[\tan^{-1}\large\frac{a}{\sqrt{b^2-a^2}}\big]$ then $x$ is equal to

$\begin{array}{1 1}(a)\;\large\frac{b}{\sqrt{2b^2-a^2}}&(b)\;\large\frac{a}{\sqrt{2b^2-a^2}}\\(c)\;\large\frac{\sqrt{2b^2-a^2}}{a}&(d)\;\large\frac{\sqrt{2b^2-a^2}}{b}\end{array}$

Given :
$\cot(\cos^{-1}x)=\sec\big[\tan^{-1}\large\frac{a}{\sqrt{b^2-a^2}}\big]$