Trigonometry! #42

Let cos(x)=4/5 . Implies tan (x) = 3/4 (using Pythagoras Theorem)

Therefore cos^-1(4/5) = tan^-1(3/4)

Since tan^-1(a) + tan^-1(b) = tan^-1 [(a+b)/1-ab]

The equation left is: tan(tan^-1[{(3/4)+(2/3)}/{1- (3/4)(2/3)}]) = 17/6

Therefore the answer is : None of these

Let $\cos x =\left( \frac{4}{5}\right) \Rightarrow \tan x = \left(\frac{3}{4}\right)$

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

Hence, the expression becomes $\large{\tan\left(\tan^{-1}\left(\frac{3}{4}\right)+\tan^{-1}\left(\frac{2}{3}\right)\right) \\ = \frac{\frac{3}{4}+\frac{2}{3}}{1-\left(\frac{3}{4}\right)\left(\frac{2}{3}\right)} \\ = \boxed{\frac{17}{6}}}$