Question

Show that : $tan\left(\begin{array}{c}\frac{1}{2}si{n}^{-1}\frac{3}{4}\end{array}\right)=\frac{4-\sqrt{7}}{3}$.

Answer 1

Let $\theta =si{n}^{-1}\frac{3}{4}$

We know that $si{n}^{-1}x=ta{n}^{-1}\frac{x}{\sqrt{1-x^2}}$

Therefore, $\theta =si{n}^{-1}\frac{3}{4}=ta{n}^{-1}\frac{\frac{3}{4}}{\sqrt{(1-\frac{3}{4}{)}^2}}$

$\theta =ta{n}^{-1}\frac{\frac{3}{4}}{\frac{\sqrt{7}}{\sqrt{16}}}$

$=ta{n}^{-1}(\frac{3}{4}\times \frac{4}{\sqrt{7}})$

$\theta =ta{n}^{-1}\frac{3}{\sqrt{7}}$

$tan\theta =\frac{3}{\sqrt{7}}$

We know that $tanx=\frac{2tan\frac{x}{2}}{1-ta{n}^2\frac{x}{2}}$

$tan\theta =\frac{2tan\frac{\theta }{2}}{1-ta{n}^2\frac{\theta }{2}}=\frac{3}{\sqrt{7}}$

$3-3ta{n}^2\frac{\theta }{2}=\frac{2}{\sqrt{7}}tan\frac{\theta }{2}$

$tan\frac{\theta }{2}=\frac{-2\sqrt{7}\pm \sqrt{28+36}}{6}$

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

Since $sin\frac{\theta }{2}$ is acute, $tan\frac{\theta }{2}$ which is the R.H.S of the expression.

Since we started with substitution $\theta =si{n}^{-1}\frac{3}{4}$ L.H.S = $tan\left(\frac{1}{2}si{n}^{-1}\frac{3}{4}\right)=\frac{4-\sqrt{7}}{3}$

$tan\frac{\theta }{2}=\frac{4-\sqrt{7}}{3}$, which is the L.H.S. of the expression.

Therefore, L.H.S = R.H.S.