If $\cot heta = \frac{7}{8}$, then the value of $\frac{(1 + \sin heta)(1 - \sin heta)}{(1 + \cos heta)(1 - \cos

Question

If $\cot 	heta = \frac{7}{8}$, then the value of $\frac{(1 + \sin 	heta)(1 - \sin 	heta)}{(1 + \cos 	heta)(1 - \cos 	heta)}$ is:

Accepted Answer

Simplifying the expression using $(1+x)(1-x) = 1-x^2$: $\frac{1 - \sin^2 	heta}{1 - \cos^2 	heta} = \frac{\cos^2 	heta}{\sin^2 	heta} = (\frac{\cos 	heta}{\sin 	heta})^2 = \cot^2 	heta$. Given $\cot 	heta = \frac{7}{8}$, so $(\frac{7}{8})^2 = \frac{49}{64}$.

Introduction to Trigonometry

❓ Sibling Questions in this Chapter