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. $\frac{b}{\sqrt{2b^2-a^2}}$
• 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

Answer: (A)

We have, $cot\left(co{s}^{-1}x\right)=sec\left(ta{n}^{-1}\frac{a}{\sqrt{b^2-a^2}}\right)$
Let $ta{n}^{-1}\frac{a}{\sqrt{b^2-a^2}}=\theta$
$\Rightarrow tan\theta =\frac{a}{\sqrt{b^2-a^2}}$
$sec\theta =\frac{b}{\sqrt{b^2-a^2}}$

$\therefore cot\left(co{s}^{-1}x\right)=\frac{b}{\sqrt{b^2-a^2}}$
$\Rightarrow co{s}^{-1}x=co{t}^{-1}\left(\frac{b}{\sqrt{b^2-a^2}}\right)$
Again, let $co{t}^{-1}\left(\frac{b}{\sqrt{b^2-a^2}}\right)=\varphi$
$\Rightarrow cot\varphi =\frac{b}{\sqrt{b^2-a^2}}$

$\Rightarrow cos\varphi =\frac{b}{\sqrt{b^2-a^2}}$
Now, $co{s}^{-1}x=\varphi$
$\Rightarrow x=cos\varphi =\frac{b}{\sqrt{2b^2-a^2}}$