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If $cot \theta=\frac{7}{8}$, evaluate :
$ \frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$
Given :
The given term is $cot \theta=\frac{7}{8}$.
To do :
We have to evaluate $ \frac{(1+sin \theta)(1-sin \theta)}{(1+cos \theta)(1-cos \theta)}$ by using $cot \theta=\frac{7}{8}$.
Solution :
$ \frac{(1+sin \theta)(1-sin \theta)}{(1+cos \theta)(1-cos \theta)}$
The Numerator and the denominator are in the form of $(a + b)(a - b)$
We know that, $(a + b)(a - b) = a^2 - b^2$
So, $ \frac{(1+sin \theta)(1-sin \theta)}{(1+cos \theta)(1-cos \theta)} = \frac{1-sin^2 \theta}{1-cos^2 \theta}$
From, $sin^2 \theta + cos^2 \theta = 1$
It can infer that, $ cos^2 \theta = 1 - sin^2 \theta ; sin^2 \theta = 1- cos^2 \theta$
$ \frac{1-sin^2 \theta}{1-cos^2 \theta} = \frac{cos^2 \theta}{sin^2 \theta}$
$\frac{cos^2 \theta}{sin^2 \theta} = cot^2 \theta$ $[\frac{cos \theta}{sin \theta} = cot \theta]$
It's given that , $cot \theta=\frac{7}{8}$
$cot^2 \theta = (\frac{7}{8})^2$
$cot^2 \theta = \frac{49}{64}$.
The value of $ \frac{(1+sin \theta)(1-sin \theta)}{(1+cos \theta)(1-cos \theta)}$ when $cot \theta=\frac{7}{8}$ is $\frac{49}{64}$