Question

Q. (1/cos ​θ - 1/ sinθ) (1 + sinθ/cosθ+ cosθ/sinθ) = tanθ secθ -cot θ cosec θ

Answer 1

Dear student

$$\begin{gathered} \left(\frac{1}{\mathrm{cos\theta }}-\frac{1}{\sin {\displaystyle \mathrm{\theta }}}\right)\left(1+\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}+\frac{\mathrm{cos\theta }}{\mathrm{sin\theta }}\right) \\ =\left(\mathrm{sec\theta }-\mathrm{cosec\theta }\right)\left(1+\mathrm{tan\theta }+\mathrm{cot\theta }\right) \\ =\mathrm{sec\theta }+\mathrm{sec\theta }\mathrm{tan\theta }+\mathrm{sec\theta }\mathrm{cot\theta }-\mathrm{cosec\theta }-\mathrm{cosec\theta }\mathrm{tan\theta }-\mathrm{cosec\theta }\mathrm{cot\theta } \\ =\mathrm{sec\theta }+\mathrm{sec\theta }\mathrm{tan\theta }+\cancel{\left(\frac{1}{\mathrm{cos\theta }}\times \frac{\mathrm{cos\theta }}{\mathrm{sin\theta }}\right)-\frac{1}{\mathrm{sin\theta }}}-\mathrm{cosec\theta }\mathrm{tan\theta }-\mathrm{cosec\theta }\mathrm{cot\theta } \\ =\cancel{\mathrm{sec\theta }-\left(\frac{1}{\mathrm{sin\theta }}\times \frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}\right)}+\mathrm{sec\theta }\mathrm{tan\theta }-\mathrm{cosec\theta }\mathrm{cot\theta } \\ =\mathrm{sec\theta }\mathrm{tan\theta }-\mathrm{cosec\theta }\mathrm{cot\theta } \\ =\mathrm{RHS} \end{gathered}$$
Regards