Question

(i) $\frac{1-{\tan }^2\mathrm{\theta }}{1+{\tan }^2\mathrm{\theta }}=({\cos }^2\mathrm{\theta }-{\sin }^2\mathrm{\theta })$
(ii) $\frac{1-{\tan }^2\mathrm{\theta }}{{\cot }^2\mathrm{\theta }-1}={\tan }^2\mathrm{\theta }$

Answer 1

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