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Mathematics Question on Inverse Trigonometric Functions

If f(x)=2tan⁡−1x+sin⁡−1(2x1+x2),x>1f\left(x\right) = 2 \tan^{-1} x + \sin^{-1} \left(\frac{2x}{1+x^{2}}\right), x > 1, then f(t)f(t) is equal to :

A

π2\frac{\pi}{2}

B

π\pi

C

4 tan⁡−1(5)4\, \tan^{-1} \left(5\right)

D

tan⁡−1(65156) \tan^{-1} \left(\frac{65}{156}\right)

Answer

π\pi

Explanation

Solution

Given that:

f(x)=2tan⁡−1x+sin⁡−1(2x1+x2)f(x) = 2\tan^{-1} x + \sin^{-1} (\frac{2x}{1+x^2}) x >0

For x>1
sin⁡−1(2x1+x2)\sin^{-1} (\frac{2x}{1+x^2}) = π−2tan⁡−1x\pi-2\tan^{-1} x

⇒ f(x) = 2tan⁡−1π+π−2tan⁡−1x2\tan^{-1}\pi + \pi-2\tan^{-1} x

⇒ f(x) = π\pi

∴f(5)=π\therefore f(5) = \pi