If $\cot \theta=\frac{7}{8},$ evaluate:

$(i)$ $\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$

$(ii)$ $\cot ^{2} \theta$

Let us consider a right triangle $ABC ,$ right-angled at point $B$.

$\cot \theta=\frac{\text { Side adjacent to } \angle \theta}{\text { Side opposite to } \angle \theta}=\frac{B C}{A B}$

$=\frac{7}{8}$

If $B C$ is $7 k,$ then $A B$ will be $8 k,$ where $k$ is a positive integer.

Applying Pythagoras theorem in $\triangle ABC ,$ we obtain

$AC ^{2}= AB ^{2}+ BC ^{2}$

$=(8\, k)^{2}+(7\, k)^{2}$

$=64\, k^{2}+49\, k^{2}$

$=113 \,k^{2}$

$A C=\sqrt{113 k}$

$\sin \theta=\frac{\text { Side opposite to } \angle \theta}{\text { Hypotenuse }}=\frac{A B}{A C}$

$=\frac{8 k}{\sqrt{113} k}=\frac{8}{\sqrt{113}}$

$\cot \theta=\frac{\text { Side adjacent to } \angle \theta}{\text { Hypotenuse }}=\frac{B C}{A C}$

$=\frac{7 k}{\sqrt{113} k}=\frac{7}{\sqrt{113}}$

$\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}=\frac{\left(1-\sin ^{2} \theta\right)}{\left(1-\cos ^{2} \theta\right)}$

$(i)$

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

$=\frac{\frac{49}{113}}{\frac{64}{113}}=\frac{49}{64}$

$(ii)$ $\cot ^{2} \theta=(\cot \theta)^{2}=\left(\frac{7}{8}\right)^{2}=\frac{49}{64}$