If $$tan\theta = \frac{7}{8}$$, then evaluate $$\frac{(1 + \sin\theta)(1 - \sin\theta)}{(1 + \cos\theta)(1 - \cos\theta)(\cot\theta)}$$.​

Correct option is A

Given:

​$$\tan\theta=\dfrac{7}{8}.$$​​

Evaluate
​$$\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)(\cot\theta)}$$​.

Formula Used:

​$$(1+\sin\theta)(1-\sin\theta)=1-\sin^2\theta=\cos^2\theta$$​​

$$(1+\cos\theta)(1-\cos\theta)=1-\cos^2\theta=\sin^2\theta$$​​

​$$\cot\theta=\dfrac{\cos\theta}{\sin\theta}$$​ and $$\cot\theta=\dfrac{1}{\tan\theta}$$​​

Solution:
​$$\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)(\cot\theta)} \\ \ \\=\frac{\cos^2\theta}{\sin^2\theta\cdot \cot\theta} \\ \ \\=\frac{\cos^2\theta}{\sin^2\theta\cdot \frac{\cos\theta}{\sin\theta}} \\ \ \\ =\frac{\cos\theta}{\sin\theta} \\ \ \\=\cot\theta .$$​​
Given $$\tan\theta=\dfrac{7}{8}\implies \cot\theta=\dfrac{8}{7}$$​