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  4. If cot (cos-1 x) = sec (tan-1 (a/√b2 - a2)), then x is equal to

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

Inverse Trigonometric Functions

Solution:

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