Question

Show that $\cot \left({\sin }^{-1}\sqrt{\frac{13}{17}}\right)=\sin \left({\tan }^{-1}\frac{2}{3}\right)$

Answer 1

$Letx={\sin }^{-1}\sqrt{\frac{13}{17}}$

$\sin x=\sqrt{\frac{13}{17}}=\frac{Altitude}{Hypotenuse}$

$Hence,Base=\sqrt{{\sqrt{17}}^2-{\sqrt{13}}^2}=\sqrt{17-13}=\sqrt{4}=2.$

$Cotx=\frac{Base}{Altitude}=\frac{2}{\sqrt{13}}$

Similarly,

$x={\tan }^{-1}\frac{2}{3}$

$\tan x=\frac{2}{3}=\frac{Altitude}{Base}$

$Hence,Hypotenuse=\sqrt{3^2+2^2}=\sqrt{9+4}=\sqrt{13}.$

$\sin x=\frac{Altitude}{Hypotenuse}=\frac{2}{\sqrt{13}}.$

Hence proved.