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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:

$cot\left(cos^{-1}\,x\right)=sec\left(tan^{-1} \frac{a}{\sqrt{b^{2}-a^{2}}}\right)\,...(i)$
Let $\theta=tan^{-1} \frac{a}{\sqrt{b^{2}-a^{2}}}$
$\therefore tan\,\theta=\frac{a}{\sqrt{b^{2}-a^{2}}}$
image
In $\Delta ABC, AC^{2}=a^{2}+\left(\sqrt{b^{2}-a^{2}}\right)^{2}$
$AC=b$
$\therefore sec \,\theta=\frac{b}{\sqrt{b^{2}-a^{2}}}\,...\left(ii\right)$
from $\left(i\right), cot\left(cos^{-1}\,x\right)=sec\,\theta$
$cot\left(cos^{-1}\,x\right)=\frac{b}{\sqrt{b^{2}-a^{2}}}\left[ using \left(ii\right)\right]$
$cos^{-1}\,x=cot^{-1}\left(\frac{b}{\sqrt{b^{2}-a^{2}}}\right)$.............(iii)
Again let $ \alpha = cot ^{-1}\left(\frac{b}{\sqrt{b^{2}-a^{2}}}\right)$
$cot\,\alpha=\frac{b}{\sqrt{b^{2}-a^{2}}}$
image
$\therefore PR^{2}=\left(\sqrt{b^{2}-a^{2}}\right)^{2}+b^{2}
\, PR= \sqrt{2b^{2}-a^{2}}
\,\,\therefore cos\,\alpha=\frac{b}{\sqrt{b^{2}-a^{2}}}\,,,,\left(iv\right)$
From $\left(iii\right),$
$cos^{-1}\,x=\alpha$ or $x=cos\,\alpha$
$x=\frac{b}{\sqrt{b^{2}-a^{2}}} \left[using \left(iv\right)\right]$