Question
If $\cot\theta=\frac{7}{8},$ evaluate:
$\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}$
$\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}$
✓
Answer
Given, $\cot\theta=\frac{7}{8}$
Now, $\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}=\frac{1-\sin^2\theta}{1-\cos^2\theta}$ $\begin{cases}1-\sin^2\theta=\cos^2\theta\\1-\cos^2\theta=\sin^2\theta\end{cases}$
$=\frac{\cos^2\theta}{\sin^2\theta}$
$=\cot^2\theta$
$=(\cot\theta)^2$
$=\Big(\frac{7}{8}\Big)^2=\frac{49}{64}$
Now, $\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}=\frac{1-\sin^2\theta}{1-\cos^2\theta}$ $\begin{cases}1-\sin^2\theta=\cos^2\theta\\1-\cos^2\theta=\sin^2\theta\end{cases}$
$=\frac{\cos^2\theta}{\sin^2\theta}$
$=\cot^2\theta$
$=(\cot\theta)^2$
$=\Big(\frac{7}{8}\Big)^2=\frac{49}{64}$
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