Question

If $cot\left(co{s}^{-1}x\right)=sec\left(ta{n}^{-1}\frac{a}{\sqrt{b^2-a^2}}\right),$ then $x$ is equal to

• A. $\frac{a}{\sqrt{2b^2-a^2}}$
• B. $\frac{b}{\sqrt{2b^2-a^2}}$
• C. $\frac{\sqrt{2b^2-a^2}}{a}$
• D. $\frac{\sqrt{2b^2-a^2}}{b}$

Answer 1

Answer: (C)

$cot\left(co{s}^{-1}x\right)=sec\left(ta{n}^{-1}\frac{a}{\sqrt{b^2-a^2}}\right)...(i)$
Let $\theta =ta{n}^{-1}\frac{a}{\sqrt{b^2-a^2}}$
$\therefore tan\theta =\frac{a}{\sqrt{b^2-a^2}}$

In $\Delta ABC,A{C}^2=a^2+{\left(\sqrt{b^2-a^2}\right)}^2$
$AC=b$
$\therefore sec\theta =\frac{b}{\sqrt{b^2-a^2}}...\left(ii\right)$
from $\left(i\right),cot\left(co{s}^{-1}x\right)=sec\theta$
$cot\left(co{s}^{-1}x\right)=\frac{b}{\sqrt{b^2-a^2}}\left[using\left(ii\right)\right]$
$co{s}^{-1}x=co{t}^{-1}\left(\frac{b}{\sqrt{b^2-a^2}}\right)$.............(iii)
Again let $\alpha =co{t}^{-1}\left(\frac{b}{\sqrt{b^2-a^2}}\right)$
$cot\alpha =\frac{b}{\sqrt{b^2-a^2}}$

$\therefore P{R}^2={\left(\sqrt{b^2-a^2}\right)}^2+b^2<br/>PR=\sqrt{2b^2-a^2}<br/>\therefore cos\alpha =\frac{b}{\sqrt{b^2-a^2}},,,\left(iv\right)$
From $\left(iii\right),$
$co{s}^{-1}x=\alpha$ or $x=cos\alpha$
$x=\frac{b}{\sqrt{b^2-a^2}}\left[using\left(iv\right)\right]$