Show that: (1 2 ^ -1 3 4 ) = 4-7 3 . - Vidyadip

Question

Show that: $\tan\bigg(\frac{1}{2}\sin^{-1}\frac{3}{4}\bigg) = \frac{4-\sqrt{7}}{3}.$

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Answer

Let $\sin^{-1}\Big(\frac{3}{4}\Big) = \theta\Rightarrow\sin\theta=\frac{3}{4}\Bigg[\theta\in\bigg(- \frac{\pi}{2},\frac{\pi}{2}\bigg)\Bigg]$
$\Rightarrow\frac{2\tan\frac{\theta}{2}}{1 + \tan^{2}\frac{\theta}{2}} = \frac{3}{4}\bigg[\because\sin2\text{x} = \frac{2\tan\text{x}}{1+ \tan^{2}\text{x}}\bigg]$
$\Rightarrow3 + 3\tan^{2}\frac{\theta}{2} = 8\tan\frac{\theta}{2}\Rightarrow3\tan^{2}\frac{\theta}{2} - 8\tan\frac{\theta}{2} + 3 = 0 $
$\Rightarrow\tan\frac{\theta}{2} = \frac{8\pm\sqrt{64 - 36}}{6}\Rightarrow\tan\frac{\theta}{2} = \frac{8\pm\sqrt{28}}{6}$
$\Rightarrow\tan\frac{\theta}{2} = \frac{8\pm2\sqrt{7}}{6}\Rightarrow\tan\frac{\theta}{2} = \frac{4\pm\sqrt{7}}{3}$
$\Rightarrow\tan\bigg(\frac{1}{2}\sin^{-1}\frac{3}{4}\bigg) = \frac{4 - \sqrt{7}}{3}\bigg[\because\theta = \sin^{-1}\frac{3}{4}\bigg].$

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