The value of $\tan \left(\cos ^{-1} \frac{3}{5}+\tan ^{-1} \frac{1}{4}\right)$
(a) $\frac{19}{8}$
(b) $\frac{8}{19}$
(C) $\frac{19}{12}$
(d) $\frac{3}{4}$
$\tan \left(\cos ^{-1} \frac{3}{5}+\tan ^{-1} \frac{1}{4}\right)=\tan \left(\tan ^{-1} \frac{\sqrt{1-\frac{9}{25}}}{\frac{3}{5}}+\tan ^{-1} \frac{1}{4}\right)$
$=\tan \left(\tan ^{-1} \frac{\frac{4}{5}}{\frac{3}{5}}+\tan ^{-1} \frac{1}{4}\right)$
$=\tan \left(\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{1}{4}\right)$
$=\tan \left(\tan ^{-1} \frac{\frac{4}{3}+\frac{1}{4}}{1-\frac{1}{3}}\right)$
$=\frac{\frac{16+3}{12}}{\frac{2}{3}}$
$=\frac{19}{8}$
Hence, the correct answer is option (a).
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.