# Evaluate the following:

Question:

Evaluate the following:

(i) $\tan \left\{2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right\}$

(ii) $\tan \frac{1}{2}\left(\cos ^{-1} \frac{\sqrt{5}}{3}\right)$

(iii) $\sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)$

(iv) $\sin \left(2 \tan ^{-1} \frac{2}{3}\right)+\cos \left(\tan ^{-1} \sqrt{3}\right)$

Solution:

(i)

$\tan \left(2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right)=\tan \left(2 \tan ^{-1} \frac{1}{5}-\tan ^{-1} 1\right)$

$=\tan \left[\tan ^{-1}\left\{\frac{{ }^{2 \times \frac{1}{5}}}{1-\left(\frac{1}{5}\right)^{2}}\right\}-\tan ^{-1} 1\right]\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left\{\frac{2 x}{1-x^{2}}\right\}\right]$

$=\tan \left[\tan ^{-1}\left\{\frac{\frac{2}{5}}{\frac{24}{25}}\right\}-\tan ^{-1} 1\right]$

$=\tan \left[\tan ^{-1} \frac{5}{12}+\tan ^{-1} 1\right]$

$\left.=\tan \left[\tan ^{-1}\left(\frac{\frac{5}{12}-1}{1+\frac{5}{12}}\right)\right]\right]\left[\because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\right]$

$=\tan \left[\tan ^{-1}\left(\frac{\frac{-7}{12}}{\frac{17}{12}}\right)\right]$

$=\tan \left[\tan ^{-1} \frac{-7}{17}\right]$

$=\frac{-7}{17}$

(ii)

Let, $\cos ^{-1} \frac{\sqrt{5}}{3}=\theta$

$\Rightarrow \cos \theta=\frac{\sqrt{5}}{3}$

$\Rightarrow 2 \cos ^{2} \frac{\theta}{2}-1=\frac{\sqrt{5}}{3}$

$\Rightarrow \cos ^{2} \frac{\theta}{2}=\frac{3+\sqrt{5}}{6}$

$\Rightarrow \cos \frac{\theta}{2}=\sqrt{\frac{3+\sqrt{5}}{6}}$

$\Rightarrow \frac{\theta}{2}=\cos ^{-1}\left(\sqrt{\frac{3+\sqrt{5}}{6}}\right)$

$=\tan ^{-1}\left(\frac{\sqrt{1-\left(\sqrt{\frac{3+\sqrt{5}}{6}}\right)^{2}}}{\sqrt{\frac{3+\sqrt{6}}{6}}}\right)$

$=\tan ^{-1}\left(\frac{\sqrt{1-\frac{3+\sqrt{6}}{6}}}{\sqrt{\frac{3+\sqrt{5}}{6}}}\right)$

$=\tan ^{-1}\left(\frac{\sqrt{\frac{3 \sqrt{6}}{6}}}{\sqrt{\frac{3+\sqrt{6}}{6}}}\right)$

$=\tan ^{-1}\left(\sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}}\right)$

$=\tan ^{-1}\left(\sqrt{\frac{(3-\sqrt{5})(3-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}}\right)$

$=\tan ^{-1}\left(\sqrt{\frac{(3-\sqrt{5})^{2}}{9-5}}\right)$

$=\tan ^{-1}\left(\frac{3-\sqrt{5}}{2}\right)$

i. e., $\frac{1}{2}\left(\cos ^{-1} \frac{\sqrt{5}}{3}\right)=\tan ^{-1}\left(\frac{3-\sqrt{5}}{2}\right)$

$\Rightarrow \tan \frac{1}{2}\left(\cos ^{-1} \frac{\sqrt{5}}{3}\right)=\tan \left[\tan ^{-1}\left(\frac{3-\sqrt{5}}{2}\right)\right]$

$\therefore \tan \frac{1}{2}\left(\cos ^{-1} \frac{\sqrt{5}}{3}\right)=\frac{3-\sqrt{5}}{2}$

(iii)

$\sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)=\sin \left\{\frac{1}{2} \times 2 \sin ^{-1} \pm \sqrt{\frac{1-\frac{4}{5}}{2}}\right\} \quad\left[\because \cos ^{-1} x=2 \sin ^{-1} \pm \sqrt{\frac{1-x}{2}}\right]$

$=\sin \left(\sin ^{-1} \pm \frac{1}{\sqrt{10}}\right)$

$=\pm \frac{1}{\sqrt{10}}$

(iv)

$\sin \left(2 \tan ^{-1} \frac{2}{3}\right)+\cos \left(\tan ^{-1} \sqrt{3}\right)=\sin \left(\sin ^{-1} \frac{2 \times \frac{2}{3}}{1+\frac{4}{9}}\right)+\cos \left(\cos ^{-1} \frac{1}{\sqrt{1+(\sqrt{3})^{2}}}\right)$

$=\sin \left(\sin ^{-1} \frac{12}{13}\right)+\cos \left(\cos ^{-1} \frac{1}{2}\right)$

$=\frac{12}{13}+\frac{1}{2}$

$=\frac{37}{26}$

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