Question

Prove the following : $\cos \left[\tan ^{-1}\left\{\sin \left(\cot ^{-1} x\right)\right\}\right]=\sqrt{\frac{1+x^2}{2+x^2}}$

Answer 1

SOLUTION : L.H.S. $=\cos \left[\tan ^{-1} \sin \left(\cot ^{-1} x\right)\right]$

Put $\cot ^{-1} \mathrm{x}=\mathrm{t} \quad \Rightarrow \quad \cot \mathrm{t}=\mathrm{x}$

$\therefore \operatorname{cosec}^2 \mathrm{t}=1+\cot ^2 \mathrm{t} \quad \Rightarrow \quad \operatorname{cosec} \mathrm{t}=\sqrt{1+\mathrm{x}^2}$

$\therefore \sin t=\frac{1}{\sqrt{1+x^2}} \quad \therefore$ L.H.S. $=\cos \left[\tan ^{-1} \sin t\right]=\cos \left[\tan ^{-1} \frac{1}{\sqrt{1+x^2}}\right]$

Let $\tan ^{-1} \frac{1}{\sqrt{1+x^2}}=z \Rightarrow \tan z=\frac{1}{\sqrt{1+x^2}} \quad \therefore \quad \sec ^2 z=1+\tan ^2 z=1+\frac{1}{1+x^2}=\frac{2+x^2}{1+x^2}$

$\sec z=\sqrt{\frac{2+x^2}{1+x^2}} \quad \Rightarrow \quad \cos z=\sqrt{\frac{1+x^2}{2+x^2}}$

Hence L.H.S. $=\cos z=\sqrt{\frac{1+x^2}{2+x^2}}=$ R.H.S.