Question

Show that each of the following is an identity:
(i) $\frac{{\tan }^2\mathrm{\theta }}{(1+{\tan }^2\mathrm{\theta })}={\sin }^2\mathrm{\theta }$
(ii) $\frac{\mathrm{cot\theta }+\mathrm{cos\theta }}{\mathrm{cot\theta }-\mathrm{cos\theta }}=\frac{1+\mathrm{sin\theta }}{1-\mathrm{sin\theta }}$

Answer 1

$$\begin{gathered} \begin{array}{l}\\ \left(\text{i}\right)\text{LHS=}\frac{{\text{tan}}^2\theta }{(1+{\text{tan}}^2\theta )}\\ \text{=}\frac{{\text{tan}}^2\theta }{{\text{sec}}^2\theta }\\ \text{=}\frac{\frac{{\text{sin}}^2\theta }{{\cos }^2\theta }}{\frac{1}{{\cos }^2\theta }}\\ {\text{=sin}}^2\theta \\ \text{=RHS}\\ \end{array} \\ \text{i.e., LHS=RHS} \\ \therefore \mathrm{T}\text{his is an identity}\text{.} \\ \begin{array}{l}\left(\text{ii}\right)\text{LHS=}\frac{\text{cot}\theta \text{+cos}\theta }{\cot \theta -\cos \theta }\\ \text{=}\frac{\frac{\cos \theta }{\sin \theta }+\cos \theta }{\frac{\cos \theta }{\sin \theta }-\cos \theta }\\ \text{=}\frac{\cos \theta +\cos \theta \sin \theta }{\cos \theta -\cos \theta \sin \theta }\\ \text{=}\frac{\cos \theta (1+\sin \theta )}{\cos \theta (1-\sin \theta )}\\ \text{=}\frac{1+\sin \theta }{1-\sin \theta }\\ \text{=RHS}\\ \mathrm{i}.\mathrm{e}.,\mathrm{LHS}=\mathrm{RHS}\end{array} \\ \therefore \mathrm{T}\text{his is an identity}\text{.} \end{gathered}$$