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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.
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1 Answer

  • $sin^{-1}x = \tan^{-1}\large \frac{x}{\sqrt (1-x^2)}$
  • $\tan x=\large \frac{2\tan\frac{x}{2}}{1-tan^2\frac{x}{2}}$
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

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