# Evaluate:

Question:

Evaluate:

(i) $\tan \left\{\cos ^{-1}\left(-\frac{7}{25}\right)\right\}$

(ii) $\operatorname{cosec}\left\{\cot ^{-1}\left(-\frac{12}{5}\right)\right\}$

(iii) $\cos \left(\tan ^{-1} \frac{3}{4}\right)$

Solution:

(i)

$\tan \left\{\cos ^{-1}\left(-\frac{7}{25}\right)\right\}=\tan \left\{\cos ^{-1}\left(\pi-\frac{7}{25}\right)\right\}$

$=-\tan \left\{\cos ^{-1}\left(\frac{7}{25}\right)\right\}$

$=-\tan \left\{\tan ^{-1}\left[\frac{\sqrt{1-\left(\frac{7}{25}\right)^{2}}}{\frac{7}{25}}\right]\right\}$

$=-\tan \left\{\tan \frac{24}{7}\right\}$

$=-\frac{24}{7}$

(ii)

$\operatorname{cosec}\left\{\cot ^{-1}\left(-\frac{12}{5}\right)\right\}=\operatorname{cosec}\left\{\cot ^{-1}\left(\pi-\frac{12}{5}\right)\right\}$

$=\operatorname{cosec}\left\{\cot ^{-1}\left(\frac{12}{5}\right)\right\}$

$=\operatorname{cosec}\left\{\sin ^{-1}\left(\frac{\frac{5}{12}}{\sqrt{1+\left(\frac{5}{12}\right)^{2}}}\right)\right\}$

$=\operatorname{cosec}\left\{\sin ^{-1}\left(\frac{5}{13}\right)\right\}$

$=\operatorname{cosec}\left\{\operatorname{cosec}^{-1}\left(\frac{13}{5}\right)\right\}$

$=\frac{13}{5}$

(iii) We have

$\cos \left(\tan ^{-1} \frac{3}{4}\right)=\cos \left[\frac{1}{2} \cos ^{-1}\left(\frac{1-\left(\frac{3}{4}\right)^{2}}{1+\left(\frac{3}{4}\right)^{2}}\right)\right] \quad\left[\because 2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\right]$

$=\cos \left[\frac{1}{2} \cos ^{-1}\left(\frac{7}{25}\right)\right]$

Let $y=\cos ^{-1}\left(\frac{7}{25}\right)$

$\quad \Rightarrow \cos y=\frac{7}{25}$

Now,

$\cos \left[\frac{1}{2} \cos ^{-1}\left(\frac{7}{25}\right)\right]=\cos \left[\frac{1}{2} y\right]$

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

$=\sqrt{\frac{\frac{7}{25}+1}{2}}$

$=\sqrt{\frac{32}{50}}$

$=\frac{4}{5}$

$\therefore \cos \left[\tan ^{-1}\left(\frac{3}{4}\right)\right]=\frac{4}{5}$