Question

Prove the following: $1+\left(\frac{{\cot }^2\alpha }{1+\cos ec\alpha }\right)=\mathrm{cosec}\alpha$

Answer 1

Proof of trigonometric identity:

$1+\left(\frac{{\cot }^2\alpha }{1+\cos ec\alpha }\right)=\mathrm{cosec}\alpha$

Taking LCM

$$\begin{gathered} \mathrm{LHS}=1+\left(\frac{{\cot }^2\alpha }{1+\cos ec\alpha }\right) \\ =1+\left(\frac{{\displaystyle \frac{{\cos }^2\alpha }{{\sin }^2\alpha }}}{1+{\displaystyle \frac{1}{\sin \alpha }}}\right)\mathbf{}\left[\because \cot \alpha =\frac{\cos \alpha }{\sin \alpha }\right] \\ =1+\frac{{\cos }^2\alpha }{{\sin }^2\alpha }·\frac{\sin \alpha }{1+\sin \alpha } \\ =\frac{{\sin }^2\alpha (1+\sin \alpha )+\sin \alpha ·{\cos }^2\alpha }{{\sin }^2\alpha (1+\sin \alpha )} \\ =\frac{{\sin }^2\alpha +{\sin }^3\alpha +\sin \alpha ·{\cos }^2\alpha }{{\sin }^2\alpha (1+\sin \alpha )} \\ =\frac{{\sin }^2\alpha +\sin \alpha ({\sin }^2\alpha +{\cos }^2\alpha )}{{\sin }^2\alpha (1+\sin \alpha )} \\ =\frac{\sin \alpha (1+\sin \alpha )}{{\sin }^2\alpha (1+\sin \alpha )}\left[\because {\sin }^2\alpha +{\cos }^2\alpha =1\right] \\ =\frac{1}{\sin \alpha } \\ =\mathrm{cosec}\alpha \left[\because \frac{1}{\sin \alpha }=\mathrm{cosec}\alpha \right] \\ =\mathrm{RHS} \end{gathered}$$

Hence, it is proved that $1+\left(\frac{{\cot }^2\alpha }{1+\cos ec\alpha }\right)=\mathrm{cosec}\alpha$