Question

Evaluate the following:

(i) $\tan \left\{2{\tan }^{-1}\frac{1}{5}-\frac{\mathrm{\pi }}{4}\right\}$
(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) $\sin \left(2{\tan }^{-1}\frac{2}{3}\right)+\cos \left({\tan }^{-1}\sqrt{3}\right)$

Answer 1

(i)

$$\begin{gathered} \tan \left(2{\tan }^{-1}\frac{1}{5}-\frac{\mathrm{\pi }}{4}\right)=\tan \left(2{\tan }^{-1}\frac{1}{5}-{\tan }^{-1}1\right) \\ =\tan \left[{\tan }^{-1}\left\{\frac{2\times {\displaystyle \frac{1}{5}}}{1-{\left({\displaystyle \frac{1}{5}}\right)}^2}\right\}-{\tan }^{-1}1\right]\left[\because 2{\tan }^{-1}x={\tan }^{-1}\left\{\frac{2x}{1-x^2}\right\}\right] \\ =\tan \left[{\tan }^{-1}\left\{\frac{{\displaystyle \frac{2}{5}}}{{\displaystyle \frac{24}{25}}}\right\}-{\tan }^{-1}1\right] \\ =\tan \left[{\tan }^{-1}\frac{5}{12}+{\tan }^{-1}1\right] \\ =\tan \left[{\tan }^{-1}\left(\frac{{\displaystyle \frac{5}{12}-}1}{1+{\displaystyle \frac{5}{12}}}\right)\right]\left[\because {\tan }^{-1}x-{\tan }^{-1}y={\tan }^{-1}\left(\frac{{\displaystyle x}-y}{1+{\displaystyle xy}}\right)\right] \\ =\tan \left[{\tan }^{-1}\left(\frac{{\displaystyle \frac{-7}{12}}}{{\displaystyle \frac{17}{12}}}\right)\right] \\ =\tan \left[{\tan }^{-1}\frac{-7}{17}\right] \\ =\frac{-7}{17} \end{gathered}$$

(ii)
$$\begin{gathered} \mathrm{Let},{\cos }^{-1}\frac{\sqrt{5}}{3}=\theta \\ \Rightarrow \cos \theta =\frac{\sqrt{5}}{3} \\ \Rightarrow 2{\cos }^2\frac{\theta }{2}-1=\frac{\sqrt{5}}{3} \\ \Rightarrow {\cos }^2\frac{\theta }{2}=\frac{3+\sqrt{5}}{6} \\ \Rightarrow \cos \frac{\theta }{2}=\sqrt{\frac{3+\sqrt{5}}{6}} \\ \Rightarrow \frac{\theta }{2}={\cos }^{-1}\left(\sqrt{\frac{3+\sqrt{5}}{6}}\right) \\ ={\tan }^{-1}\left(\frac{\sqrt{1-{\left(\sqrt{\frac{3+\sqrt{5}}{6}}\right)}^2}}{\sqrt{\frac{3+\sqrt{5}}{6}}}\right) \\ ={\tan }^{-1}\left(\frac{\sqrt{1-\frac{3+\sqrt{5}}{6}}}{\sqrt{\frac{3+\sqrt{5}}{6}}}\right) \\ ={\tan }^{-1}\left(\frac{\sqrt{\frac{3-\sqrt{5}}{6}}}{\sqrt{\frac{3+\sqrt{5}}{6}}}\right) \\ ={\tan }^{-1}\left(\sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}}\right) \\ ={\tan }^{-1}\left(\sqrt{\frac{\left(3-\sqrt{5}\right)\left(3-\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}}\right) \\ ={\tan }^{-1}\left(\sqrt{\frac{{\left(3-\sqrt{5}\right)}^2}{9-5}}\right) \\ ={\tan }^{-1}\left(\frac{3-\sqrt{5}}{2}\right) \\ \mathrm{i}.\mathrm{e}.,\frac{1}{2}\left({\cos }^{-1}\frac{\sqrt{5}}{3}\right)={\tan }^{-1}\left(\frac{3-\sqrt{5}}{2}\right) \\ \Rightarrow \tan \frac{1}{2}\left({\cos }^{-1}\frac{\sqrt{5}}{3}\right)=\tan \left[{\tan }^{-1}\left(\frac{3-\sqrt{5}}{2}\right)\right] \\ \therefore \tan \frac{1}{2}\left({\cos }^{-1}\frac{\sqrt{5}}{3}\right)=\frac{3-\sqrt{5}}{2} \end{gathered}$$

(iii)
$$\begin{gathered} \sin \left(\frac{1}{2}{\cos }^{-1}\frac{4}{5}\right)=\sin \left\{\frac{1}{2}\times 2{\sin }^{-1}\pm \sqrt{\frac{1-\frac{4}{5}}{2}}\right\}\left[\because {\cos }^{-1}x=2{\sin }^{-1}\pm \sqrt{\frac{1-x}{2}}\right] \\ =\sin \left({\sin }^{-1}\pm \frac{1}{\sqrt{10}}\right) \\ =\pm \frac{1}{\sqrt{10}} \end{gathered}$$

(iv)
$$\begin{gathered} \sin \left(2{\tan }^{-1}\frac{2}{3}\right)+\cos \left({\tan }^{-1}\sqrt{3}\right)=\sin \left({\sin }^{-1}\frac{2\times {\displaystyle \frac{2}{3}}}{1+{\displaystyle \frac{4}{9}}}\right)+\cos \left({\cos }^{-1}\frac{1}{\sqrt{1+{\left(\sqrt{3}\right)}^2}}\right) \\ =\sin \left({\sin }^{-1}\frac{12}{13}\right)+\cos \left({\cos }^{-1}\frac{1}{2}\right) \\ =\frac{12}{13}+\frac{1}{2} \\ =\frac{37}{26} \end{gathered}$$