Browse Questions

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

(Note: This question has been split into 2 questions) This question appeared in 65-1,65-2 and 65-3 versions of the paper in 2013.

Toolbox:
• $sin^{-1}x = \tan^{-1}\large \frac{x}{\sqrt (1-x^2)}$
• $\tan x=\large \frac{2\tan\frac{x}{2}}{1-tan^2\frac{x}{2}}$
Given $tan\bigg(\frac{1}{2}sin^{-1}\frac{3}{4}\bigg)=\frac{4-\sqrt 7}{3}$.
$\textbf{Step 1}$:
Let $\theta = sin^{-1}\frac{3}{4}$
We know that $sin^{-1}x = \tan^{-1}\large \frac{x}{\sqrt (1-x^2)}$
Therefore, $\theta = sin^{-1}\frac{3}{4} = \tan^{-1}\large \frac{\large \frac{3}{4}}{\sqrt (1-(\large \frac{3}{4})^2)}$
$\Rightarrow \theta $$= \tan^{-1}\large \frac{\large \frac{3}{4}}{\sqrt \large\frac{16-9}{16}}$$ = tan^{-1} (\frac{3}{4} \times \frac{\sqrt16}{\sqrt7})$
$\Rightarrow \theta = tan^{-1} \frac{3}{\sqrt7}$
$\Rightarrow tan \theta = \frac{3}{\sqrt7}$
$\textbf{Step 2}$:
We know that $\tan x=\large \frac{2\tan\frac{x}{2}}{1-tan^2\frac{x}{2}}$