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Inverse Trigonometric Functions — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions5 Marks
Question
Prove that $\cos \left[\tan ^{-1}\left\{\sin \left(\cot ^{-1} x\right)\right\}\right]=\sqrt{\frac{1+x^2}{2+x^2}}$
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Answer

$\text { L.H.S. }=\cos \left[\tan ^{-1} \sin \left(\cot ^{-1} x\right)\right]$
Suppose $\cot ^{-1} x=t \Rightarrow \cot t=x$
$\therefore \operatorname{cosec}^2 t=1+\cot ^2 t$
$\operatorname{cosec} t=\sqrt{1+x^2}$
$\therefore \sin t=\frac{1}{\sqrt{1+x^2}}$
$\therefore \text { L.H.S. }=\cos \left[\tan ^{-1} \sin t\right]$
$=\cos \left[\tan ^{-1} \frac{1}{\sqrt{1+x^2}}\right]$
suppose $\tan ^{-1} \frac{1}{\sqrt{1+x^2}}=Z \Rightarrow \tan Z=\frac{1}{\sqrt{1+x^2}}$
$\therefore \sec ^2 Z=1+\tan ^2 Z$
$=1+\frac{1}{1+x^2}=\frac{1+x^2+1}{1+x^2}$
$=\frac{2+x^2}{1+x^2}$
$\therefore \sec Z=\sqrt{\frac{2+x^2}{1+x^2}} \Rightarrow \cos Z=\sqrt{\frac{1+x^2}{2+x^2}}$
So, $\text { L.H.S. }=\cos Z$
$=\sqrt{\frac{1+x^2}{2+x^2}}=\text { R.H.S. }$

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