Question

$if \: \: \sin^{ - 1} x - \cos ^{ - 1} = \frac{\pi}{6} \\ then \: find \: the \: value \: of \: x$
Chapter: Inverse Trigonometry

Answer: √3/2

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Answer 1

The answer is given below :

$\sin^{ - 1} x - \cos ^{ - 1}x = \frac{\pi}{6} \: \: ...(1)$

We know that

$\sin^{ - 1} x + \cos ^{ - 1} x= \frac{\pi}{2} \: \: ...(2)$

Now, adding (1) and (2), we get

$2 {sin}^{ - 1} x = \frac{\pi}{6} + \frac{\pi}{2} \\ \\ Or, \: \: 2 {sin}^{ - 1} x = \frac{\pi + 3\pi}{6} \\ \\ Or ,\: \: 2 {sin}^{ - 1} x = \frac{4\pi}{6} \\ \\ Or, \: \: {sin}^{ - 1} x = \frac{4\pi}{12 } \\ \\ Or, \: \: {sin}^{ - 1} x = \frac{\pi}{3} \\ \\ Or, \: \: x = sin \frac{\pi}{3} \\ \\ Or, \: \: x = \frac{ \sqrt{3} }{2}$

So, the value of x is = √3/2.

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