Question

If $cot\left({\cos }^{-1}x\right)=sec\left[{\tan }^{-1}\left(\frac{a}{\sqrt{b^2-a^2}}\right)\right]$, then $x=$

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

Answer 1

The correct option is A

$\frac{b}{\sqrt{b^2–a^2}}$

Step 1. Find the value of $x$:

Given, $cot\left({\cos }^{-1}x\right)=sec\left[{\tan }^{-1}\left(\frac{a}{\sqrt{b^2-a^2}}\right)\right]$

Let $\theta ={\tan }^{-1}\frac{a}{\sqrt{b^2–a^2}}$

$\Rightarrow$$sec\theta =\frac{b}{\sqrt{b^2–a^2}}$

Step 2. Put the value of $\theta$ in given equation, we get

$cot\left({\cos }^{-1}x\right)=sec\theta$

$\Rightarrow$$cot\left({\cos }^{-1}x\right)=\frac{b}{\sqrt{b^2–a^2}}$

Let $ɑ=co{t}^{-1}\left[\frac{b}{\sqrt{b^2–a^2}}\right]$

$\Rightarrow$ $cotɑ=\frac{b}{\sqrt{b^2–a^2}}$

$\Rightarrow$${\cos }^{-1}x=ɑ$

$\Rightarrow$ $x=\cos ɑ$

$\mathbf{\therefore }\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{b}}{\sqrt{{\mathbf{b}}^{\mathbf{2}}\mathbf{}\mathbf{–}\mathbf{}{\mathbf{a}}^{\mathbf{2}}}}$

Hence, Option ‘A’ is Correct.