Solution :
(i) In \(\triangle\) ABC, \(cot \theta\) = \(7\over 8\) = \(AB\over BC\)
Let AB = 7k and BC = 8k
Now, AC = \(\sqrt{{AB}^2 + {BC}^2}\) = \(\sqrt{113k^2}\)
So, AC = \(\sqrt{113}k\)
Thus, \(sin \theta\) = \(8k\over \sqrt{113}k\) = \(8\over \sqrt{113}\)
\(cos \theta\) = \(7k\over \sqrt{113}k\) = \(7\over \sqrt{113}\)
Now, \({(1 + sin\theta)(1 – sin\theta)}\over {(1 + cos\theta)(1 – cos\theta)}\) = \(49\over 64\)
(ii) \(cot^2 \theta\) = \(({7\over 8})^2\) = \(49\over 64\)