Question

Prove that $\frac{{\cot }^2\theta \left(\sec \theta -1\right)}{\left(1+\sin \theta \right)}+\frac{{\sec }^2\theta \left(\sin \theta -1\right)}{\left(1+\sec \theta \right)}=0$

Answer 1

$$\begin{gathered} \mathrm{LHS}=\frac{{\cot }^2\theta \left(\sec \theta -1\right)}{\left(1+\sin \theta \right)}+\frac{{\sec }^2\theta \left(\sin \theta -1\right)}{\left(1+\sec \theta \right)} \\ =\frac{{\displaystyle \frac{{\cos }^2\theta }{{\sin }^2\theta }}\left({\displaystyle \frac{1}{\cos \theta }}-1\right)}{\left(1+\sin \theta \right)}+\frac{{\displaystyle \frac{1}{{\cos }^2\theta }}\left(\sin \theta -1\right)}{\left(1+{\displaystyle \frac{1}{\cos \theta }}\right)} \\ =\frac{{\displaystyle \frac{{\cos }^2\theta }{{\sin }^2\theta }\left(\frac{1-\cos \theta }{\cos \theta }\right)}}{\left(1+\sin \theta \right)}+\frac{{\displaystyle \frac{\left(\sin \theta -1\right)}{{\cos }^2\theta }}}{\left({\displaystyle \frac{\cos \theta +1}{\cos \theta }}\right)} \\ =\frac{{\displaystyle {\cos }^2\theta \left(1-\cos \theta \right)}}{{\sin }^2\theta \cos \theta \left(1+\sin \theta \right)}+\frac{{\displaystyle \left(\sin \theta -1\right)\cos \theta }}{\left({\displaystyle \cos \theta +1}\right){\cos }^2\theta } \\ =\frac{{\displaystyle \cos \theta \left(1-\cos \theta \right)}}{\left(1-{\cos }^2\theta \right)\left(1+\sin \theta \right)}+\frac{{\displaystyle \left(\sin \theta -1\right)\cos \theta }}{\left({\displaystyle \cos \theta +1}\right)\left(1-{\sin }^2\theta \right)} \\ =\frac{{\displaystyle \cos \theta \left(1-\cos \theta \right)}}{\left(1-\cos \theta \right)\left(1+\cos \theta \right)\left(1+\sin \theta \right)}+\frac{{\displaystyle -\left(1-\sin \theta \right)\cos \theta }}{\left({\displaystyle \cos \theta +1}\right)\left(1-\sin \theta \right)\left(1+\sin \theta \right)} \\ =\frac{\cos \theta }{\left(1+\cos \theta \right)\left(1+\sin \theta \right)}-\frac{\cos \theta }{\left(\cos \theta +1\right)\left(1+\sin \theta \right)} \\ =0 \\ =\mathrm{RHS} \end{gathered}$$