Question : If $\cos 21^{\circ}=\frac{x}{y}$, then $(\operatorname{cosec21^{\circ}}-\cos 69^{\circ})$ is equal to:
Option 1: $\frac{x^{2}}{y\sqrt{y^{2}-x^{2}}}$
Option 2:
$\frac{y^{2}}{x\sqrt{y^{2}-x^{2}}}$
Option 3:
$\frac{y^{2}}{x\sqrt{x^{2}-y^{2}}}$
Option 4: $\frac{x^{2}}{y\sqrt{x^{2}-y^{2}}}$
Correct Answer: $\frac{x^{2}}{y\sqrt{y^{2}-x^{2}}}$
Solution :
Given:
$\cos 21^{\circ}=\frac{x}{y}$
So, $\cos 69^{\circ}=\cos (90^{\circ}-21^{\circ})=\sin 21^{\circ}= \sqrt{1-\cos^2 21º}=\sqrt{1-\frac{x^2}{y^2}}=\frac{\sqrt{y^2-{x^2}}}{y}$
Also, $\operatorname{cosec21^{\circ}}=\frac{1}{\sin 21^{\circ}}=\frac{y}{\sqrt{y^2-{x^2}}}$
Now, $(\operatorname{cosec21^{\circ}}-\cos 69^{\circ})=\frac{y}{\sqrt{y^2-{x^2}}}-\frac{\sqrt{y^2-{x^2}}}{y}$
⇒ $(\operatorname{cosec21^{\circ}}-\cos 69^{\circ})=\frac{y^2-(y^2-x^2)}{y\sqrt{y^2-{x^2}}}=\frac{x^{2}}{y\sqrt{y^{2}-x^{2}}}$
Hence, the correct answer is $\frac{x^{2}}{y\sqrt{y^{2}-x^{2}}}$.
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