Question

show that tan[1/2sin inverse(3/4)]=4-√7/3​

Answer 1

Answer:

I think this method is most simple for you.

Answer 2

$\huge \bold \color{lime}{ \underline{ \color{teal} \underline \mathbb{ANSWER }}}$

To prove :-

$\tan( \frac{1}{2} { \sin}^{ - 1} \frac{3}{4} ) = \frac{4 - \sqrt{7} }{3}$

Proof :-

$let \: \frac{1}{2} { \sin }^{ - 1} \frac{3}{4} = x$

$2x = { \sin }^{ - 1} \frac{3}{4}$

$\sin (2x) = \frac{3}{4}$

$\frac{2 \tan(x) }{1 + { \tan }^{2} x} = \frac{3}{4}$

$8 \tan(x) = 3 + 3 { \tan }^{2} x$

$3 { \tan }^{2} x - 8 \tan(x) + 3 = 0$

$\tan(x) = \frac{8 \pm \: \sqrt{64 - 36} }{6}$

$\tan(x) = \frac{8 \pm \: \sqrt{28} }{6}$

$\tan(x) = \frac{8 \pm \: 2 \sqrt{7} }{6}$

$\tan(x) = \frac{4 \pm \: \sqrt{7} }{3}$

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

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

$\small \tan( - \frac{\pi}{4} ) \leqslant \tan( \frac{1}{2} { \sin}^{ - 1} \frac{3}{4} ) \leqslant \tan( \frac{\pi}{4} )$

$- 1 \leqslant \tan( \frac{1}{2} { \sin }^{ - 1} \frac{3}{4} ) \leqslant 1$

$\small \bold{ \therefore \: \tan(x) = \tan( \frac{1}{2} { \sin }^{ - 1} \frac{3}{4} ) = \huge\color{green} \frac{4 - \sqrt{7} }{3} }$

$\mathcal{HOPE \: IT \: HELPS }$