Question

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

Answer 1

Given $\cot \theta =\frac{7}{8}$
Consider $\frac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}$$=\frac{1-{\sin }^2\theta }{1-{\cos }^2\theta }$
$=\frac{co{s}^2\theta }{si{n}^2\theta }$
[$\because 1-{\sin }^2\theta ={\cos }^2\theta$ and $1-{\cos }^2\theta ={\sin }^2\theta$ ]
$\Rightarrow \frac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}=(\frac{cos\theta }{sin\theta }{)}^2$
$=(\cot \theta {)}^2$
$=(\frac{7}{8}{)}^2$ [ Given value]
$=\frac{49}{64}$
Therefore, the value of the given expression is $\frac{49}{64}$

Alternatively ,

Let $\Delta ABC$ in which $\angle B={90}^{\circ }$ and $\angle C=\theta$
According to the question,
$\cot \theta =\frac{BC}{AB}=\frac{7}{8}$
Let $BC=7k$ and $AB=8k$, where $k$ is a positive real number.
By Pythagoras theorem in $\Delta ABC$; we get,
$A{C}^2=A{B}^2+B{C}^2$
$A{C}^2=(8k{)}^2+(7k{)}^2$
$A{C}^2=64k^2+49k^2$
$A{C}^2=113k^2$
$AC=\sqrt{113}k$
$\sin \theta =\frac{AB}{AC}=\frac{8k}{\sqrt{113}k}=\frac{8}{\sqrt{113}}$
and $\cos \theta =\frac{BC}{AC}=\frac{7k}{\sqrt{113}k}=\frac{7}{\sqrt{113}}\dots \dots \left(i\right)$
$\Rightarrow$$\frac{(1+\sin \theta )(1-sin\theta )}{(1+\cos \theta )(1-\cos \theta )}=\frac{(1-{\sin }^2\theta )}{(1-{\cos }^2\theta )}$

$=\frac{\{1-{\left(\frac{8}{\sqrt{113}}\right)}^2\}}{\{1-{\left(\frac{7}{\sqrt{113}}\right)}^2\}}=\frac{\{1-\left(\frac{64}{113}\right)\}}{\{1-\left(\frac{49}{113}\right)\}}=\frac{\left\{\frac{(113-64)}{113}\right\}}{\left\{\frac{(113-49)}{113}\right\}}=\frac{49}{64}$