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Mathematics Question on Inverse Trigonometric Functions

If 2sin−1x−3cos−1x=42sin^{-1}x - 3cos^{-1}x = 4, then 2sin−1x+3cos−1x2sin^{-1}x + 3cos^{-1}x is equal to

A

6π−45\frac{6\pi -4}{5}

B

4−6π5\frac{4 - 6\pi}{5}

C

3π2\frac{3\pi }{2}

D

00

Answer

6π−45\frac{6\pi -4}{5}

Explanation

Solution

2sin−1x−3cos−1x=42sin^{-1}x - 3cos^{-1}x = 4 ⇒2(π2−cos−1 x)−3cos−1x=4\Rightarrow 2\left(\frac{\pi}{2}-cos^{-1}\,x\right) - 3cos^{-1}x = 4 ⇒π−5cos−x=4\Rightarrow \pi - 5cos^{-x} = 4 ⇒cos−1x=π−45\Rightarrow cos^{-1}x =\frac{\pi - 4}{5} ⇒3cos−1x=3π−125⋯(i)\Rightarrow 3cos^{-1}x = \frac{3\pi -12}{5}\quad\cdots\left(i\right) Similarly, 2sin−1x−3(π2−sin−1 x)=42sin^{-1}x - 3\left(\frac{\pi}{2}-sin^{-1}\,x\right) = 4 ⇒5sin−1x=4+3π2\Rightarrow 5sin^{-1}x = 4 + \frac{3\pi}{2} ⇒sin−1x=(45+3π2)\Rightarrow sin^{-1}x = \left(\frac{4}{5}+\frac{3\pi }{2}\right) ⇒2sin−1x=(85+3π5)⋯(ii)\Rightarrow 2sin^{-1}x = \left(\frac{8}{5}+\frac{3\pi }{5}\right)\quad\cdots\left(ii\right) Adding e (i)\left(i\right) and (ii)\left(ii\right), we get 2sin−1x+3cos−1x=8+3π5+3π−1252 sin^{-1} x + 3 cos^{-1}x = \frac{8+3\pi}{5}+\frac{3\pi-12}{5} =6π−45= \frac{6\pi -4}{5}