Show that tan-1 (1/2 sin-1 3/4) = (4 – √7)/ 3

We have, tan-1 (1/2 sin-1 3/4)

Let ½ sin-1 ¾ = θ ⇒ sin-1 ¾ = 2θ ⇒ sin 2θ = ¾

2 tan θ/ 1 + tan2 θ = ¾

3 tan2 θ – 8 tan θ + 3 = 0

$\tan \theta=\frac{8 \pm \sqrt{64-36}}{6}$

$\tan \theta=\frac{8 \pm \sqrt{28}}{6}=\frac{8 \pm 2 \sqrt{7}}{6}=\frac{4 \pm \sqrt{7}}{3}$

Now,

$-\frac{\pi}{2} \leq \sin ^{-1} \frac{3}{4} \leq \frac{\pi}{2}$

$\frac{-\pi}{4} \leq \frac{1}{2} \sin ^{-1} \frac{3}{4} \leq \frac{\pi}{2}$

$\tan \left(\frac{-\pi}{4}\right) \leq \tan \left(\frac{1}{2}\left(\sin ^{-1} \frac{3}{4}\right)\right) \leq \tan \frac{\pi}{4}$

$-1 \leq \tan \left(\frac{1}{2} \sin ^{-1} \frac{3}{4}\right) \leq 1$

Hence,

$\tan \theta=\frac{4-\sqrt{7}}{3}\left(\tan \theta=\frac{4+\sqrt{7}}{3}>1\right.$, which is not possible $)$