If $Cot \theta =\frac{7}{8}$, evaluate : (i) $\frac{\left(1+\sin \theta \right)\left(1-\sin \theta \right)}{\left(1+\cos \theta \right)\left(1-\cos \theta \right)}$  (ii) $co{t}^2\theta$

Let we have a triangle, 

$Cot \theta =\frac{7}{8}$

Since $Cot \left(\theta \right) =\frac{Adjacent side}{Opposite side}=\frac{AB}{BC}=\frac{7}{8}$

Let AB be 7k and BC will be 8k

where k is a positive real number.

Using Pythagoras theorem, we get

$A{C}^2=A{B}^2 + B{C}^2$

${\left(AC\right)}^2={\left(7k\right)}^2 + {\left(8k\right)}^2$

$A{C}^2=49k^2+64k^2$

$A{C}^2=113k^2$

Thus, $AC =\sqrt{113}k$

$Sin \theta =\frac{Opposite side}{Hypotenuse}= \frac{BC}{AC}=\frac{8k}{\sqrt{113}k}=\frac{8}{\sqrt{113}}$

$Cos \theta =\frac{Adjacent side}{Hypotenuse}= \frac{AB}{AC}= \frac{7k}{\sqrt{113}k}=\frac{7}{\sqrt{113}}$

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

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

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