Question

If $2\sin x=1,\frac{\pi }{2}<x<\pi$ and $\sqrt{2}\cos y=1,\frac{3\pi }{2}<y<2\pi ,$ find the value of $\frac{\tan x+\tan y}{\cos x-\cos y}$.

Answer 1

We have,

$2\sin x=1$

$\sin x=\frac{1}{2}$

$\sin x=\sin {30}^o$

$x={30}^o$

If $\sqrt{2}\cos y=1$

$\cos y=\frac{1}{\sqrt{2}}$

$\cos y=\cos {45}^o$

$y={45}^o$

Then, value of

$\frac{\tan x+\tan y}{\cos x-\cos y}$

$=\frac{\tan {30}^o+\tan {45}^o}{\cos {30}^o-\cos {45}^o}$

$=\frac{\frac{1}{\sqrt{3}}+1}{\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}}$

$=\frac{\frac{1+\sqrt{3}}{\sqrt{3}}}{\frac{\sqrt{6}-2}{2\sqrt{2}}}$

$=\frac{2\sqrt{2}(1+\sqrt{3})}{\sqrt{3}(\sqrt{6}-2)}$

$=\frac{2\sqrt{2}(1+\sqrt{3})}{\sqrt{2}\sqrt{3}(\sqrt{3}-\sqrt{2})}$

$=\frac{2(1+\sqrt{3})}{\sqrt{3}(\sqrt{3}-\sqrt{2})}$

On rationalize and we get,

$=\frac{2(1+\sqrt{3})}{\sqrt{3}(\sqrt{3}-\sqrt{2})}\times \frac{(\sqrt{3}+\sqrt{2})}{(\sqrt{3}+\sqrt{2})}\times \frac{\sqrt{3}}{\sqrt{3}}$

$=\frac{2\sqrt{3}(1+\sqrt{3})(\sqrt{3}+\sqrt{2})}{3[{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2]}$

$=\frac{2\sqrt{3}(1+\sqrt{3})(\sqrt{3}+\sqrt{2})}{3[3-2]}$

$=\frac{2\sqrt{3}(1+\sqrt{3})(\sqrt{3}+\sqrt{2})}{3}$.