Question

PT: cos(90 -θ) / 1+sin(90-θ) + 1+sin(90-θ)/cos(90-θ) = 2cosecθ

Answer 1

$$\begin{gathered} \mathbf{LHS}=\frac{\cos \left(90°-\mathrm{\theta }\right)}{1+\sin \left(90°-\mathrm{\theta }\right)}+\frac{1+\sin \left(90°-\mathrm{\theta }\right)}{\cos \left(90°-\mathrm{\theta }\right)} \\ =\frac{\sin \mathrm{\theta }}{1+\cos \mathrm{\theta }}+\frac{1+\cos \mathrm{\theta }}{\sin \mathrm{\theta }} \\ =\frac{{\sin }^2\mathrm{\theta }+{\left(1+\cos \mathrm{\theta }\right)}^2}{\left(1+\cos \mathrm{\theta }\right)\sin \mathrm{\theta }} \\ =\frac{{\sin }^2\mathrm{\theta }+1+{\cos }^2\mathrm{\theta }+2\cos \mathrm{\theta }}{\left(1+\cos \mathrm{\theta }\right)\sin \mathrm{\theta }} \\ =\frac{\left({\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }\right)+1+2\cos \mathrm{\theta }}{\left(1+\cos \mathrm{\theta }\right)\sin \mathrm{\theta }} \\ =\frac{1+1+2\cos \mathrm{\theta }}{\left(1+\cos \mathrm{\theta }\right)\sin \mathrm{\theta }} \\ =\frac{2+2\cos \mathrm{\theta }}{\left(1+\cos \mathrm{\theta }\right)\sin \mathrm{\theta }} \\ =\frac{2\left(1+\cos \mathrm{\theta }\right)}{\left(1+\cos \mathrm{\theta }\right)\sin \mathrm{\theta }} \\ =\frac{2}{\sin \mathrm{\theta }} \\ =2\mathrm{cosec}\mathrm{\theta } \\ =\mathbf{RHS} \end{gathered}$$