Question

prove that

1+cosec θ+cotθ /1+cosecθ - cotθ = 1+ cosθ = 1+cosθ /sinθ

Answer 1

$$\begin{gathered} \mathrm{LHS}=\frac{\mathrm{cosec}\mathrm{\theta }+\cot \mathrm{\theta }+1}{1+\mathrm{cosec}\mathrm{\theta }-\cot \mathrm{\theta }} \\ =\frac{\mathrm{cosec}\mathrm{\theta }+\cot \mathrm{\theta }+\left({\mathrm{cosec}}^2\mathrm{\theta }-{\cot }^2\mathrm{\theta }\right)}{1+\mathrm{cosec}\mathrm{\theta }-\cot \mathrm{\theta }} \\ =\frac{\left(\mathrm{cosec}\mathrm{\theta }+\cot \mathrm{\theta }\right)+\left(\mathrm{cosec}\mathrm{\theta }+\cot \mathrm{\theta }\right)\left(\mathrm{cosec}\mathrm{\theta }-\cot \mathrm{\theta }\right)}{\left(1+\mathrm{cosec}\mathrm{\theta }-\cot \mathrm{\theta }\right)} \\ =\frac{\left(\mathrm{cosec}\mathrm{\theta }+\cot \mathrm{\theta }\right)\left(1+\mathrm{cosec}\mathrm{\theta }-\cot \mathrm{\theta }\right)}{\left(1+\mathrm{cosec}\mathrm{\theta }-\cot \mathrm{\theta }\right)} \\ =\left(\mathrm{cosec}\mathrm{\theta }+\cot \mathrm{\theta }\right) \\ =\frac{1}{\sin \mathrm{\theta }}+\frac{\cos \mathrm{\theta }}{\sin \mathrm{\theta }} \\ =\frac{1+\cos \mathrm{\theta }}{\sin \mathrm{\theta }} \\ =\mathrm{RHS} \end{gathered}$$