Solve for $x$ : $\sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}$ - Noon Academy

Solve for $x$ : $\sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}$

CBSE | Class 12 | Exercise 4C | Inverse Trigonometric Functions | Maths | RS Aggarwal

Solve for $x$ : $\sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}$

Solution:

Given: $\sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}$
We know that $\sin ^{-1} \mathrm{x}+\cos ^{-1} \mathrm{x}=\frac{\pi}{2}$
So, $\sin ^{-1} x=\frac{\pi}{2}-\cos ^{-1} x$
Substituting in the given equation,
$\frac{\pi}{2}-\cos ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}$
Rearranging,
$\begin{array}{l} 2 \cos ^{-1} x=\frac{\pi}{2}-\frac{\pi}{6} \\ 2 \cos ^{-1} x=\frac{\pi}{3} \\ \cos ^{-1} x=\frac{\pi}{6} \\ x=\frac{\sqrt{3}}{2} \end{array}$
Therefore, $\mathrm{x}=\frac{\sqrt{3}}{2}$ is the required value of $\mathrm{x}$ .

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