Question

(i) sec θ (1 − sin θ) (sec θ + tan θ) = 1
(ii) sin θ(1 + tan θ) + cos θ(1 + cot θ) = (sec θ + cosec θ)

Answer 1

$\begin{array}{l}\\ \left(\text{i}\right)\text{LHS=sec}\theta (1-\text{sin}\theta )(\text{sec}\theta +\text{tan}\theta )\\ \text{=}(\text{sec}\theta -\text{sec}\theta \text{sin}\theta )(\text{sec}\theta +\text{tan}\theta )\\ \text{=}(\text{sec}\theta -\frac{1}{\text{cos}\theta }\times \text{sin}\theta )(\text{sec}\theta +\text{tan}\theta )\\ \text{=}(\text{sec}\theta -\text{tan}\theta )(\text{sec}\theta +\text{tan}\theta )\\ {\text{=sec}}^2\theta -{\text{tan}}^2\theta \\ \text{=1}\\ \text{=RHS}\\ \left(\text{ii}\right)\text{LHS=sin}\theta (1+\text{tan}\theta )+\text{cos}\theta (1+\text{cot}\theta )\\ \text{=sin}\theta +\text{sin}\theta \text{×}\frac{\text{sin}\theta }{\text{cos}\theta }+\text{cos}\theta \text{+cos}\theta \text{×}\frac{\text{cos}\theta }{\text{sin}\theta }\\ \text{=}\frac{\cos \theta {\text{sin}}^2\theta +{\text{sin}}^3\theta +{\text{cos}}^2\theta \text{sin}\theta {\text{+cos}}^3\theta }{\text{cos}\theta \text{sin}\theta }\\ \text{=}\frac{({\text{sin}}^3\theta +{\text{cos}}^3\theta )+(\text{cos}\theta {\text{sin}}^2\theta +{\text{cos}}^2\theta \text{sin}\theta )}{\text{cos}\theta \text{sin}\theta }\\ \text{=}\frac{(\text{sin}\theta +\text{cos}\theta )({\text{sin}}^2\theta -\text{sin}\theta \text{cos}\theta {\text{+cos}}^2\theta )+\text{sin}\theta \text{cos}\theta \left(\text{sin}\theta \text{+cos}\theta \right)}{\text{cos}\theta \text{sin}\theta }\\ \text{=}\frac{\left(\text{sin}\theta \text{+cos}\theta \right)({\text{sin}}^2\theta +{\text{cos}}^2\theta -\text{sin}\theta \text{cos}\theta +\text{sin}\theta \text{cos}\theta )}{\text{cos}\theta \text{sin}\theta }\\ \text{=}\frac{\left(\text{sin}\theta \text{+cos}\theta \right)\left(1\right)}{\text{cos}\theta \text{sin}\theta }\\ \text{=}\frac{\text{sin}\theta }{\text{cos}\theta \text{sin}\theta }+\frac{\text{cos}\theta }{\text{cos}\theta \text{sin}\theta }\\ \text{=}\frac{1}{\text{cos}\theta }+\frac{1}{\text{sin}\theta }\\ \text{=sec}\theta \text{+cosec}\theta \\ \text{=RHS}\end{array}$