If cot θ=78, evaluate :

(i) Given: $\cot \theta=\frac{7}{8}$

To evaluate: $\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$

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

We know the following formula

$(a+b)(a-b)=a^{2}-b^{2}$

By applying the above formula in the numerator of equation (1) ,

We get,

$(1+\sin \theta)(1-\sin \theta)=1^{2}-\sin ^{2} \theta \ldots($ Where $a=1$ and $b=\sin \theta)$

$(1+\sin \theta)(1-\sin \theta)=1-\sin ^{2} \theta$...(2)

Similarly,

By applying formula $(a+b)(a-b)=a^{2}-b^{2}$ in the denominator of equation (1),

We get,

$(1+\cos \theta)(1-\cos \theta)=1^{2}-\cos ^{2} \theta \ldots$ (Where $a=1$ and $\left.b=\cos \theta\right)$

$(1+\cos \theta)(1-\cos \theta)=1-\cos ^{2} \theta$...(3)

Substituting the value of numerator and denominator of equation (1) ,from equation (2) and(3)

Therefore,

$\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}=\frac{1-\sin ^{2} \theta}{1-\cos ^{2} \theta}$....(4)

Since,

$\cos ^{2} \theta+\sin ^{2} \theta=1$

Therefore,

$\cos ^{2} \theta=1-\sin ^{2} \theta$

Also, $\sin ^{2} \theta=1-\cos ^{2} \theta$

Putting the value of $1-\sin ^{2} \theta$ and $1-\cos ^{2} \theta$ in Equation (4)

We get,

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

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

We know that, $\frac{\cos \theta}{\sin \theta}=\cot \theta$

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

Since, It is given that $\cot \theta=\frac{7}{8}$

Therefore,

$\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}=\left(\frac{7}{8}\right)^{2}$

$\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}=\frac{7^{2}}{8^{2}}$

$\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}=\frac{49}{64}$

Answer: $\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}=\frac{49}{64}$

(ii) Given: $\cot \theta=\frac{7}{8}$

To evaluate: $\cot ^{2} \theta$

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

Squaring on both sides,

We get,

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

$\cot ^{2} \theta=\frac{7^{2}}{8^{2}}$

$\cot ^{2} \theta=\frac{49}{64}$

Answer: $\cot ^{2} \theta=\frac{49}{64}$