Given that $\cos heta=\frac{\sqrt{3}}{2}$, then the value of $\frac{\operatorname{cosec}^{2} het

Question

Given that $\cos 	heta=\frac{\sqrt{3}}{2}$, then the value of $\frac{\operatorname{cosec}^{2} 	heta-\sec ^{2} 	heta}{\operatorname{cosec}^{2} 	heta+\sec ^{2} 	heta}$ is\

Professor PC Thomas Classes · Answer

(c): Given, $\cos 	heta=\frac{\sqrt{3}}{2}=\frac{B}{H}$ Let $B=\sqrt{3} k$ and $H=2 k$\ $	herefore \quad P=\sqrt{(2 k)^{2}-(\sqrt{3} k)^{2}}$\ [By Pythagoras Theorem]\ $\Rightarrow P=\sqrt{k^{2}}=k$\ $	herefore \quad \operatorname{cosec} 	heta=\frac{H}{P}=\frac{2 k}{k}=2$ and $\sec 	heta=\frac{H}{B}=\frac{2 k}{\sqrt{3 k}}=\frac{2}{\sqrt{3}}$\ $	herefore \quad \frac{\operatorname{cosec}^{2} 	heta-\sec ^{2} 	heta}{\operatorname{cosec}^{2} 	heta+\sec ^{2} 	heta}=\frac{(2)^{2}-\left(\frac{2}{\sqrt{3}}ight)^{2}}{(2)^{2}+\left(\frac{2}{\sqrt{3}}ight)^{2}}=\frac{4-\frac{4}{3}}{4+\frac{4}{3}}=\frac{8}{16}=\frac{1}{2}$\