Question
Evaluate the following:
$\Big(\frac{\sin49^\circ}{\cos41^\circ}\Big)^2+\Big(\frac{\cos41^\circ}{\sin49^\circ}\Big)^2$
$\Big(\frac{\sin49^\circ}{\cos41^\circ}\Big)^2+\Big(\frac{\cos41^\circ}{\sin49^\circ}\Big)^2$
✓
Answer
We have to find: $\Big(\frac{\sin49^\circ}{\cos41^\circ}\Big)^2+\Big(\frac{\cos41^\circ}{\sin49^\circ}\Big)^2$
Since $\sin(90^\circ-\theta)=\cos\theta$ and $\cos(90^\circ-\theta)=\sin\theta$
So, $\Big(\frac{\sin(90^\circ-41^\circ)}{\cos41^\circ}\Big)^2+\Big(\frac{\cos(90^\circ-49^\circ)}{\sin49^\circ}\Big)^2=\Big(\frac{\cos41^\circ}{\cos41^\circ}\Big)^2+\Big(\frac{\sin49^\circ}{\sin49^\circ}\Big)^2$
$= 1 + 1$
$= 2$
So value of $=\Big(\frac{\sin49^\circ}{\cos41^\circ}\Big)^2+\Big(\frac{\cos41^\circ}{\sin49^\circ}\Big)^2\text{ is }2$
Since $\sin(90^\circ-\theta)=\cos\theta$ and $\cos(90^\circ-\theta)=\sin\theta$
So, $\Big(\frac{\sin(90^\circ-41^\circ)}{\cos41^\circ}\Big)^2+\Big(\frac{\cos(90^\circ-49^\circ)}{\sin49^\circ}\Big)^2=\Big(\frac{\cos41^\circ}{\cos41^\circ}\Big)^2+\Big(\frac{\sin49^\circ}{\sin49^\circ}\Big)^2$
$= 1 + 1$
$= 2$
So value of $=\Big(\frac{\sin49^\circ}{\cos41^\circ}\Big)^2+\Big(\frac{\cos41^\circ}{\sin49^\circ}\Big)^2\text{ is }2$
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