Prove that

Solution:

(i)

$\tan ^{-1}\left(\frac{1-x^{2}}{2 x}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\frac{\pi}{2}$

$\mathrm{LHS}=\tan ^{-1}\left(\frac{1-x^{2}}{2 x}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)$

$=\tan ^{-1}\left(\frac{1-x^{2}}{2 x}\right)+\frac{\pi}{2}-\tan ^{-1}\left(\frac{1-x^{2}}{2 x}\right) \quad\left[\because \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\right]$

$=\frac{\pi}{2}=\mathrm{RHS}$

(ii)

$\sin \left\{\tan ^{-1}\left(\frac{1-x^{2}}{2 x}\right)+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\right\}=1$

$\mathrm{LHS}=\sin \left\{\tan ^{-1}\left(\frac{1-x^{2}}{2 x}\right)+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\right\}$

$=\sin \left\{\sin ^{-1}\left(\frac{\frac{1-x^{2}}{2 x}}{\sqrt{1+\frac{1-x^{2}}{2 x}}}\right)+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\right\} \quad\left[\because \tan ^{-1} x=\sin ^{-1} \frac{x}{\sqrt{1+x^{2}}}\right]$

$=\sin \left\{\sin ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\right\}$

$=\sin \left\{\frac{\pi}{2}\right\}$                    $\left[\because \tan ^{-1} x=\sin ^{-1} \frac{x}{\sqrt{1+x^{2}}}\right]$

$=1=$ RHS

(iii)
To prove:

$\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=2 \cos ^{-1} x, \frac{1}{\sqrt{2}} \leq x \leq 1$

Let us consider $\cos ^{-1}(x)=\theta$

$\Rightarrow x=\cos \theta$

Taking R.H.S.

$\sin ^{-1}\left(2 \cos \theta \sqrt{1-\cos ^{2} \theta}\right)$

$\Rightarrow \sin ^{-1}(2 \cos \theta \sin \theta)$

$\Rightarrow \sin ^{-1}(\sin 2 \theta) \quad(\because 2 \sin \theta \cos \theta=\sin 2 \theta)$

$\Rightarrow 2 \theta$

$\Rightarrow 2 \cos ^{-1} x=$ RHS $\left(\because \theta=\cos ^{-1} x\right)$

Hence, proved.