# 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 $\frac{49}{64}$

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Updated on: 10-Oct-2022

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