The value of tan

Question:

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) $\frac{3}{4}$

Solution:

$\tan \left(\cos ^{-1} \frac{3}{5}+\tan ^{-1} \frac{1}{4}\right)=\tan \left(\tan ^{-1} \frac{\sqrt{1-\frac{9}{25}}}{\frac{3}{5}}+\tan ^{-1} \frac{1}{4}\right)$

$=\tan \left(\tan ^{-1} \frac{\frac{4}{5}}{\frac{3}{5}}+\tan ^{-1} \frac{1}{4}\right)$

$=\tan \left(\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{1}{4}\right)$

$=\tan \left(\tan ^{-1} \frac{\frac{4}{3}+\frac{1}{4}}{1-\frac{1}{3}}\right)$

$=\frac{\frac{16+3}{12}}{\frac{2}{3}}$

$=\frac{19}{8}$

Hence, the correct answer is option (a).

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