If cot $$\theta$$ = $$7\over 8$$, evaluate : (i) $${(1 + sin\theta)(1 – sin\theta)}\over {(1 + cos\theta)(1 – cos\theta)}$$ (ii) $$cot^2 \theta$$

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)  $$cot^2 \theta$$ = $$({7\over 8})^2$$ = $$49\over 64$$