Addition of two inverse trigonometric functions

Question

Find the value of $$\tan(3\tan^{-1}(3)) + \cos(3\cos^{-1}\frac{1}{3}) + 1$$ The options are: $$ (a)11 \\(b) \frac{4}{27} \\(c) \frac{9}{13} \\(d) \frac{295}{351} $$

The right answer is provided as (d)

My attempt : I tried to apply the conventional formula $$\tan^{-1}a+\tan^{-1}b = \pi + \tan^{-1}\left(\frac{ab}{a+b}\right)$$ where $ab>1$, but that is making the calculation very complex. Since it's a question meant to be solved in competitive exams, can we have a quicker method to solve this question. Any help will be appreciated.

Answer 1

Using the expressions $$\tan(3x)=\frac{3\tan(x)-\tan^3(x)}{1-3\tan^2(x)}$$ $$\cos(3x)=4\cos^3(x)-3\cos(x)$$ we have that $$\tan(3\tan^{-1}(3))= \frac{3\tan(\tan^{-1}(3))-\tan^3(\tan^{-1}(3))}{1-3\tan^2(\tan^{-1}(3))}=\frac{9-27}{1-27}=\frac{9}{13}$$ $$\cos(3\cos^{-1}(1/3))=4\cos^3(\cos^{-1}(1/3))-3\cos(\cos^{-1}(1/3))=\frac{4}{27}-1$$ Thus, $$\tan(3\tan^{-1}(3))+\cos(3\cos^{-1}(1/3))+1=\frac{295}{351}$$

Answer 2

Using the trigonometric identities $$\tan (3 \theta) = \frac{3 \tan \theta - \tan^3 \theta}{1- 3 \tan^2 \theta}$$ $$\cos (3 \theta) = 4 \cos^3 \theta - 3 \cos \theta$$ you get $$\tan (3 \tan^{-1}(3))= \frac{3 \cdot 3-3^3}{1-3\cdot 3^2}=\frac{9}{13}$$ and $$\cos(3 \cos^{-1}(1/3)) = 4 \left( \frac{1}{3}\right)^3- 3 \cdot \frac{1}{3}=-\frac{23}{27}$$ then you have to sum these two numbers and you are done.