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Mathematics Question on Inverse Trigonometric Functions

5 cos⁡−1(1−x21+x2)+7 sin⁡−1(2x1+x2)−4 tan⁡−1(2x1+x2)−tan⁡−1x=5π5\,\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) +7\,\sin^{-1}\left(\frac{2x}{1+x^2}\right)-4\,\tan^{-1}\left(\frac{2x}{1+x^2}\right)-\tan^{-1}x=5\pi , then x is equal to

A

33

B

−3-\sqrt3

C

3\sqrt3

D

2

Answer

3\sqrt3

Explanation

Solution

Given that 5 cos−1(1−x21+x2)+7 sin−1(2x1+x2)5 \,cos^{-1} \left(\frac{1-x^{2}}{1+x^{2}}\right)+7\,sin^{-1}\left(\frac{2x}{1+x^{2}}\right) −4 tan−1(2x1+x2)−tan−1x=5π-4\,tan^{-1} \left(\frac{2x}{1+x^{2}}\right)-tan^{-1}x = 5\pi ∴5(2 tan−1x)+7(2 tan−1x)−4(2 tan−1x)−tan−1x=5π\therefore 5\left(2\,tan^{-1} x\right)+7\left(2\,tan^{-1}x\right)-4\left(2\,tan^{-1}x\right)- tan^{-1}x = 5\pi ∴5(2 tan−1x)+7(2 tan−1)−4(2 tan−1x)−tan−1x=5π\therefore 5\left(2\,tan^{-1}x\right)+7\left(2\,tan^{-1}\right)-4\left(2\,tan^{-1}x\right)-tan^{-1}x = 5\pi ⇒15 tan−1x=5π \Rightarrow 15\,tan^{-1}x = 5\pi ⇒tan−1x=π3\Rightarrow tan^{-1}x = \frac{\pi}{3} ⇒x=tanπ3 \Rightarrow x= tan \frac{\pi}{3} ⇒x=3 \Rightarrow x= \sqrt{3}