solve the equation for an exact solution. sin⁻¹x - tan⁻¹√3 = -π/4 choose the correct answer below. a. {(-√2…

Explanation:

Step1: Evaluate $\tan^{-1}\sqrt{3}$

We know that $\tan^{-1}\sqrt{3}=\frac{\pi}{3}$ since $\tan\frac{\pi}{3}=\sqrt{3}$. So the equation $\sin^{-1}x - \tan^{-1}\sqrt{3}=-\frac{\pi}{4}$ becomes $\sin^{-1}x-\frac{\pi}{3}=-\frac{\pi}{4}$.

Step2: Isolate $\sin^{-1}x$

Add $\frac{\pi}{3}$ to both sides of the equation: $\sin^{-1}x=-\frac{\pi}{4}+\frac{\pi}{3}$.

Step3: Calculate the right - hand side

Find a common denominator, which is 12. Then $-\frac{\pi}{4}+\frac{\pi}{3}=\frac{- 3\pi + 4\pi}{12}=\frac{\pi}{12}$. So $\sin^{-1}x=\frac{\pi}{12}$.

Step4: Solve for $x$

Take the sine of both sides: $x = \sin\frac{\pi}{12}$. Using the half - angle formula $\sin\frac{\alpha}{2}=\sqrt{\frac{1 - \cos\alpha}{2}}$, for $\alpha=\frac{\pi}{6}$, we have $\sin\frac{\pi}{12}=\sqrt{\frac{1-\cos\frac{\pi}{6}}{2}}=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}=\frac{\sqrt{2 - \sqrt{3}}}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}$.

Answer:

A. $\left{\frac{-\sqrt{2}+\sqrt{6}}{4}\right}$