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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Find the value of $\tan 22^{\large\circ}30'$

$\begin{array}{1 1}(A)\;\large\frac{1}{\sqrt 2+1}&(B)\;\large\frac{1}{\sqrt 2-1}\\(C)\;\large\frac{1}{\sqrt 3+1}&(D)\;\large\frac{1}{\sqrt 3-1}\end{array} $

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  • $\sin 2\theta=2\sin \theta\cos\theta$
  • $\cos^2\theta=\large\frac{1}{2}$$(1+\cos 2\theta)$
Given : $\tan 22^{\large\circ}30'$
Let $\theta=45^{\large\circ}$
$\therefore \large\frac{\theta}{2}$$=22^{\large\circ}30'$
$\tan \large\frac{\theta}{2}=\frac{\sin \Large\frac{\theta}{2}}{\cos \Large\frac{\theta}{2}}$
Multiply both numerator and denominator by $2\cos \large\frac{\theta}{2}$
$\Rightarrow \large\frac{2\sin \Large\frac{\theta}{2}\normalsize \sin \Large\frac{\theta}{2}}{2\cos^2\Large\frac{\theta}{2}}$
$\Rightarrow \large\frac{\sin \theta}{2\big[\large\frac{1}{2}(1+\cos^2\Large\frac{\theta}{2})\big]}$
$\Rightarrow \large\frac{\sin \theta}{1+\cos \theta}$
Where $\theta=45^{\large\circ}$
$\Rightarrow \large\frac{\sin 45^{\large\circ}}{1+\cos 45^{\large\circ}}$
$\Rightarrow \large\frac{\Large\frac{1}{\sqrt 2}}{1+\Large\frac{1}{\sqrt 2}}$
$\Rightarrow \large\frac{1}{\sqrt 2+1}$
Hence (A) is the correct answer.
answered Apr 18, 2014 by sreemathi.v

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