Question
Evaluate the following:
$\Big(\frac{\sin27^\circ}{\cos63^\circ}\Big)^2-\Big(\frac{\cos63^\circ}{\sin27^\circ}\Big)^2$

Answer

We have to find: $\Big(\frac{\sin27^\circ}{\cos63^\circ}\Big)^2-\Big(\frac{\cos63^\circ}{\sin27^\circ}\Big)^2$$$
Since $\sin(90^\circ-\theta)=\cos\theta$ and $\cos(90^\circ-\theta)=\sin\theta$
So, $\Big(\frac{\sin27^\circ}{\cos63^\circ}\Big)^2-\Big(\frac{\cos63^\circ}{\sin27^\circ}\Big)^2=\bigg[\frac{\sin(90^\circ-63^\circ)}{\cos63^\circ}\bigg]^2-\bigg[\frac{\cos(90^\circ-27^\circ)}{\sin27^\circ}\bigg]^2$
$=\Big(\frac{\cos63^\circ}{\cos63^\circ}\Big)^2-\Big(\frac{\sin27^\circ}{\sin27^\circ}\Big)^2$
$= 1 - 1$
$= 0$
So value of $\Big(\frac{\sin27^\circ}{\cos63^\circ}\Big)^2-\Big(\frac{\cos63^\circ}{\sin27^\circ}\Big)^2\text{ is }0$

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