# Question #835cb

Feb 12, 2017

Given

$\tan \frac{a}{\tan} b = \sqrt{3}$

$\implies \cot \frac{b}{\cot} a = \sqrt{3}$

$\implies \cot b = \sqrt{3} \cot a \ldots \ldots \left[1\right]$

Again

$\sin \frac{a}{\sin} b = \sqrt{2}$

$\implies \csc \frac{b}{\csc} a = \sqrt{2}$

$\implies {\csc}^{2} \frac{b}{\csc} ^ 2 a = 2$

$\implies \frac{1 + {\cot}^{2} b}{1 + {\cot}^{2} a} = 2$

Putting $\cot b = \sqrt{3} \cot a$

$\implies \frac{1 + 3 {\cot}^{2} a}{1 + {\cot}^{2} a} = 2$

$\implies 1 + 3 {\cot}^{2} a = 2 + 2 {\cot}^{2} a$

$\implies 3 {\cot}^{2} a - 2 {\cot}^{2} a = 2 - 1 = 1$

$\implies \cot a = 1 = \cot \left(\frac{\pi}{4}\right) = \cot \left(\pi + \frac{\pi}{4}\right)$

$\implies a = \frac{\pi}{4} \mathmr{and} \frac{5 \pi}{4}$

Inserting $a = \frac{\pi}{4} \mathmr{and} \frac{5 \pi}{4}$ in [1]

$\implies \cot b = \sqrt{3} \cot a$

$\implies \cot b = \sqrt{3} \times 1 = \cot \left(\frac{\pi}{6}\right) = \cot \left(\pi + \frac{\pi}{6}\right)$

$\implies b = \frac{\pi}{6} \mathmr{and} \frac{7 \pi}{6}$