If 0

Question

(A) Given that $\alpha = \sin^{-1} x + \cos^{-1} \left( \frac{x}{2} + \frac{\sqrt{3}}{2} \sqrt{1 - x^2} \right)$ for $0 < x < \frac{1}{2}$.Let $x = \s

Vedclass · Accepted Answer

$\frac{4}{\sqrt{3}}$

Vedclass · Answer

(A) Given that $\alpha = \sin^{-1} x + \cos^{-1} \left( \frac{x}{2} + \frac{\sqrt{3}}{2} \sqrt{1 - x^2} ight)$ for $0 Let $x = \sin 	heta$. Since $0 Then $\sqrt{1 - x^2} = \cos 	heta$.Substituting these into the expression for $\alpha$:$\alpha = \sin^{-1}(\sin 	heta) + \cos^{-1} \left( \frac{1}{2} \sin 	heta + \frac{\sqrt{3}}{2} \cos 	heta ight)$.Using the identity $\cos(A - B) = \cos A \cos B + \sin A \sin B$,we have $\frac{1}{2} \sin 	heta + \frac{\sqrt{3}}{2} \cos 	heta = \sin \frac{\pi}{6} \sin 	heta + \cos \frac{\pi}{6} \cos 	heta = \cos \left( 	heta - \frac{\pi}{6} ight)$.Since $0 Thus,$\cos^{-1} \left( \cos \left( 	heta - \frac{\pi}{6} ight) ight) = \cos^{-1} \left( \cos \left( \frac{\pi}{6} - 	heta ight) ight) = \frac{\pi}{6} - 	heta$.Therefore,$\alpha = 	heta + \frac{\pi}{6} - 	heta = \frac{\pi}{6}$.Finally,$	an \alpha + \cot \alpha = 	an \frac{\pi}{6} + \cot \frac{\pi}{6} = \frac{1}{\sqrt{3}} + \sqrt{3} = \frac{1 + 3}{\sqrt{3}} = \frac{4}{\sqrt{3}}$.

If $0 < x < \frac{1}{2}$ and $\alpha = \sin^{-1} x + \cos^{-1} \left( \frac{x}{2} + \frac{\sqrt{3 - 3 x^2}}{2} \right)$,then $\tan \alpha + \cot \alpha =$

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