Question

(i) (1 + cos θ) (1 − cos θ) (1 + cot²θ) = 1
(ii) cosec θ (1 + cos θ) (cosec θ − cot θ) = 1

Answer 1

$\begin{array}{l}\\ \left(\text{i}\right)\text{LHS=}(1+\text{cos}\theta )(1-\text{cos}\theta )(1+{\text{cot}}^2\theta )\\ \text{=}(1-{\text{cos}}^2\theta ){\text{cosec}}^2\theta \\ {\text{=sin}}^2\theta \times {\text{cosec}}^2\theta \\ {\text{=sin}}^2\theta \times \frac{1}{{\text{sin}}^2\theta }\\ \text{=1}\\ \text{=RHS}\\ \left(\text{ii}\right)\text{LHS=cosec}\theta (1+\text{cos}\theta )(\text{cosec}\theta -\text{cot}\theta )\\ \text{=}\left(\text{cosec}\theta \text{+cosec}\theta \text{×cos}\theta \right)(\text{cosec}\theta -\text{cot}\theta )\\ \text{=}(\text{cosec}\theta \text{+}\frac{1}{\text{sin}\theta }\times \text{cos}\theta )(\text{cosec}\theta -\text{cot}\theta )\\ \text{=}\left(\text{cosec}\theta \text{+cot}\theta \right)(\text{cosec}\theta -\text{cot}\theta )\\ {\text{=cosec}}^2\theta -{\text{cot}}^2\theta (\because {\text{cosec}}^2\theta -{\text{cot}}^2\theta =1)\\ \text{=1}\\ \text{=RHS}\end{array}$