Question

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

$\frac{b}{\sqrt{2b^{2} - a^{2}}}$

Answer 2

We have, $cot \left(cos^{-1}\,x\right) = sec\left(tan^{-1} \frac{a}{\sqrt{b^{2}-a^{2}}}\right)$ Let $tan^{-1} \frac{a}{\sqrt{b^{2}-a^{2}}} = \theta$ $\Rightarrow tan\,\theta = \frac{a}{\sqrt{b^{2}-a^{2}}}$ $sec\,\theta =\frac{b}{\sqrt{b^{2}-a^{2}}}$ $\therefore cot \left(cos^{-1} \,x\right) = \frac{b}{\sqrt{b^{2}-a^{2}}}$ $\Rightarrow cos^{-1}\,x = cot^{-1} \left(\frac{b}{\sqrt{b^{2}-a^{2}}}\right)$ Again, let $cot^{-1} \left(\frac{b}{\sqrt{b^{2}-a^{2}}}\right) = \phi$ $\Rightarrow cot\,\phi = \frac{b}{\sqrt{b^{2}-a^{2}}}$ $\Rightarrow cos\,\phi = \frac{b}{\sqrt{b^{2}-a^{2}}}$ Now, $cos^{-1}\, x = \phi$ $\Rightarrow x = cos\,\phi = \frac{b}{\sqrt{2b^{2}-a^{2}}}$