Question

(i) (1 − cos²θ) cosec²θ = 1
(ii) (1 + cot²θ) sin²θ = 1

Answer 1

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