Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Model Papers
0 votes

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.
Can you answer this question?

1 Answer

0 votes
  • $sin^{-1}x = \tan^{-1}\large \frac{x}{\sqrt (1-x^2)}$
  • $\tan x=\large \frac{2\tan\frac{x}{2}}{1-tan^2\frac{x}{2}}$
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 $\tan x=\large \frac{2\tan\frac{x}{2}}{1-tan^2\frac{x}{2}}$
$\Rightarrow tan \theta =\large \frac{2\tan\frac{\theta}{2}}{1-tan^2\frac{\theta}{2}}$$= \frac{3}{\sqrt 7}$
$\Rightarrow 3-3\tan^2\frac{\theta}{2}=\frac{2}{\sqrt 7} \tan\frac{\theta}{2}$
$\Rightarrow$ $3\tan^2\frac{\theta}{2}+2\sqrt 7\tan\frac{\theta}{2}-3=0$
$\Rightarrow$ $\tan\frac{\theta}{2}=\large \frac{-2\sqrt 7\pm\sqrt(28+36)}{6}$
$\Rightarrow$ $\tan\frac{\theta}{2}=\large \frac{-2\sqrt 7\pm 8}{6} = \large \frac{-\sqrt 7\pm 4}{3}$
Since $sin\frac{\theta}{2}$ is acute, $tan\frac{\theta}{2}=\frac{4-\sqrt 7}{3}$ which is the R.H.S. of the expression.
Siince we started with substitution $\theta = sin^{-1}\frac{3}{4} \rightarrow$ L.H.S. $= tan (\frac{1}{2}sin^{-1}\frac{3}{4}) =\frac{4-\sqrt 7}{3}$
$\Rightarrow tan \frac{\theta}{2} =\frac{4-\sqrt 7}{3}$, which is the L.H.S. of the expression.
Therefore, L.H.S. = R.H.S.
answered Mar 22, 2013 by balaji.thirumalai

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App