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

Solution:

In figure,



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

$\Rightarrow \quad \frac{\mathbf{A B}}{\mathbf{B C}}=\frac{\mathbf{7}}{\mathbf{8}}$

$\Rightarrow \quad \mathrm{AB}=7 \mathrm{k}$

and $B C=8 \mathrm{k}$

Now, $A C^{2}=A B^{2}+B C^{2}=(7 k)^{2}+(8 k)^{2}$

$=113 \mathrm{k}^{2}$

$\Rightarrow \quad \mathrm{AC}=\sqrt{\mathbf{1 1 3}} \mathbf{k}$

Then $\sin \theta=\frac{\mathbf{B C}}{\mathbf{A C}}=\frac{\mathbf{8 k}}{\sqrt{\mathbf{1 1 3 k}}}=\frac{\mathbf{8}}{\sqrt{\mathbf{1 1 3}}}$

and $\cos \theta=\frac{\mathbf{A B}}{\mathbf{A C}}=\frac{\mathbf{7 k}}{\sqrt{\mathbf{1 1 3}} \mathbf{k}}=\frac{\mathbf{7}}{\sqrt{\mathbf{1 1 3}}}$.

(i) $\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}=\frac{\left(1+\frac{8}{\sqrt{113}}\right)\left(1-\frac{8}{\sqrt{113}}\right)}{\left(1+\frac{7}{\sqrt{113}}\right)\left(1-\frac{7}{\sqrt{113}}\right)}$

$\frac{(\sqrt{113}+8)(\sqrt{113}-8)}{(\sqrt{113}+7)(\sqrt{113}-7)}=\frac{(\sqrt{113})^{2}-(8)^{2}}{(\sqrt{113})^{2}-(7)^{2}}$

$\left\{\because(a+b)(a-b)=a^{2}-b^{2}\right\}$

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

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