Question

Evaluate:
(i) $\tan \left\{{\cos }^{-1}\left(-\frac{7}{25}\right)\right\}$
(ii) $\mathrm{cosec}\left\{{\cot }^{-1}\left(-\frac{12}{5}\right)\right\}$
(iii) $\cos \left({\tan }^{-1}\frac{3}{4}\right)$

Answer 1

(i)
$$\begin{gathered} \tan \left\{{\cos }^{-1}\left(-\frac{7}{25}\right)\right\}=\tan \left\{{\cos }^{-1}\left(\mathrm{\pi }-\frac{7}{25}\right)\right\} \\ =-\tan \left\{{\cos }^{-1}\left(\frac{7}{25}\right)\right\} \\ =-\tan \left\{{\tan }^{-1}\left[\frac{\sqrt{1-{\left({\displaystyle \frac{7}{25}}\right)}^2}}{{\displaystyle \frac{7}{25}}}\right]\right\} \\ =-\tan \left\{\tan \frac{24}{7}\right\} \\ =-\frac{24}{7} \end{gathered}$$
(ii)
$$\begin{gathered} \mathrm{cosec}\left\{{\cot }^{-1}\left(-\frac{12}{5}\right)\right\}=\mathrm{cosec}\left\{{\cot }^{-1}\left(\mathrm{\pi }-\frac{12}{5}\right)\right\} \\ =\mathrm{cosec}\left\{{\cot }^{-1}\left(\frac{12}{5}\right)\right\} \\ =\mathrm{cosec}\left\{{\sin }^{-1}\left(\frac{{\displaystyle \frac{5}{12}}}{\sqrt{1+{\left({\displaystyle \frac{5}{12}}\right)}^2}}\right)\right\} \\ =\mathrm{cosec}\left\{{\sin }^{-1}\left(\frac{{\displaystyle 5}}{13}\right)\right\} \\ =\mathrm{cosec}\left\{{\mathrm{cosec}}^{-1}\left(\frac{{\displaystyle 13}}{5}\right)\right\} \\ =\frac{{\displaystyle 13}}{5} \end{gathered}$$
(iii) We have

$$\begin{gathered} \cos \left({\tan }^{-1}\frac{3}{4}\right)=\cos \left[\frac{1}{2}{\cos }^{-1}\left(\frac{1-{\left({\displaystyle \frac{3}{4}}\right)}^2}{1+{\left(\frac{3}{4}\right)}^2}\right)\right]\left[\because 2{\tan }^{-1}x={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] \\ =\cos \left[\frac{1}{2}{\cos }^{-1}\left(\frac{7}{25}\right)\right] \end{gathered}$$
Let $$\begin{gathered} y={\cos }^{-1}\left(\frac{7}{25}\right) \\ \Rightarrow \cos y=\frac{7}{25} \end{gathered}$$
Now,

$$\begin{gathered} \cos \left[\frac{1}{2}{\cos }^{-1}\left(\frac{7}{25}\right)\right]=\cos \left[\frac{1}{2}y\right] \\ =\sqrt{\frac{\cos y+1}{2}}\left[\because \cos 2x=2{\cos }^{2}x-1\right] \\ =\sqrt{\frac{{\displaystyle \frac{7}{25}}+1}{2}} \\ =\sqrt{\frac{32}{50}} \\ =\frac{4}{5} \end{gathered}$$
$\therefore \cos \left[{\tan }^{-1}\left(\frac{3}{4}\right)\right]=\frac{4}{5}$