If $\cot \theta=\frac{7}{8}$, evaluate :
(i) $\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$
(ii) $\cot ^{2} \theta$
(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: $\cot ^{2} \theta=\frac{49}{64}$
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