# Find the value of $\tan 22^{\large\circ}30'$

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

Toolbox:
• $\sin 2\theta=2\sin \theta\cos\theta$
• $\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.