Mark the tick against the correct answer in the following:

To Find: The value of $\tan \left\{\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right\}$

Let $x=\cos ^{-1} \frac{4}{5}$

$\Rightarrow \cos x=\frac{4}{5}=\frac{\text { adjacent side }}{\text { hypotenuse }}$

By pythagorus theroem,

(Hypotenuse ) $^{2}=$ (opposite side $)^{2}+(\text { adjacent side })^{2}$

Therefore, opposite side $=3$

$\Rightarrow \tan \mathrm{x}=\frac{\text { oppositeside }}{\text { adjacent side }}=\frac{3}{4}$

$\Rightarrow \mathrm{x}=\tan ^{-1} \frac{3}{4}$

Now $\tan \left\{\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right\}=\tan \left\{\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{2}{3}\right\}$

Since we know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$

$\tan \left\{\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{2}{3}\right\}=\tan \left(\tan ^{-1}\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\left(\frac{3}{4} \times \frac{2}{3}\right)}\right)\right)$

$=\tan \left(\tan ^{-1}\left(\frac{17}{6}\right)\right)$

$=\frac{17}{6}$