Question

Prove that $\frac{\mathrm{sin\theta }}{1-\mathrm{cos\theta }}+\frac{\mathrm{tan\theta }}{1+\mathrm{cos\theta }}=\mathrm{sec\theta }.\mathrm{cosec\theta }+\mathrm{cot\theta }$

Answer 1

As we have $\frac{\mathrm{sin\theta }}{1-\mathrm{cos\theta }}+\frac{\mathrm{tan\theta }}{1+\mathrm{cos\theta }}=\mathrm{sec\theta }.\mathrm{cosec\theta }+\mathrm{cot\theta }$

Let us start with LHS

$$\begin{gathered} \frac{\mathrm{sin\theta }}{(1–\mathrm{cos\theta })}+\frac{\mathrm{tan\theta }}{(1+\mathrm{cos\theta })}\left[\because (1–\mathrm{cos\theta })(1+\mathrm{cos\theta })=1^2-{\cos }^2\mathrm{\theta }\right] \\ =\frac{\mathrm{sin\theta }(1+\mathrm{cos\theta })+\mathrm{tan\theta }(1-\mathrm{cos\theta })}{(1–c\mathrm{os}²\mathrm{\theta })} \\ =\frac{\mathrm{sin\theta }(1+\mathrm{cos\theta })+\mathrm{tan\theta }(1-\mathrm{cos\theta })}{\left({\sin }^2\mathrm{\theta }\right)}\left[\because {\sin }^2\mathrm{\theta }=1^2-{\cos }^2\mathrm{\theta }\right] \\ =\frac{\mathrm{sin\theta }(1+\mathrm{cos\theta })+(\mathrm{tan\theta }-\mathrm{sin\theta })}{\left({\sin }^2\mathrm{\theta }\right)} \\ =\frac{\mathrm{sin\theta }\left(1+\mathrm{cos\theta }+{\displaystyle \frac{1}{\mathrm{cos\theta }}}–1\right)}{{\sin }^2\mathrm{\theta }} \\ =\frac{\left(\mathrm{cos\theta }+{\displaystyle \frac{1}{\mathrm{cos\theta }}}\right)}{\mathrm{sin\theta }} \\ =\frac{1}{\mathrm{cos\theta Sin\theta }}+\frac{\mathrm{cos\theta }}{\mathrm{sin\theta }} \\ =\mathrm{sec\theta }.\mathrm{cosec\theta }+\mathrm{cot\theta }=\mathrm{LHS} \end{gathered}$$

Hence Proved