Question

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

${(\cos ec\theta -cot\theta )}^2=\frac{1-\cos \theta }{1+\cos \theta }$

Answer 1

Proof:

Consider the LHS ,

$$\begin{gathered} \mathrm{LHS}={(\cos ec\theta -cot\theta )}^2 \\ ={\mathrm{cosec}}^2\theta +{\cot }^2\theta -2·\mathrm{cosec}\theta ·\cot \theta \left[\mathrm{Use}{(a-b)}^2=a^2+b^2-2ab\right] \\ =\frac{1}{{\sin }^2\theta }+\frac{{\cos }^2\theta }{{\sin }^2\theta }-2·\frac{1}{\sin \theta }·\frac{\cos \theta }{\sin \theta }\left[\because \cos ec\theta =\frac{1}{\sin \theta }\mathrm{and}\cot \theta =\frac{\cos \theta }{\sin \theta }\right] \\ =\frac{1+{\cos }^2\theta -2\cos \theta }{{\sin }^2\theta } \\ =\frac{{(1-\cos \theta )}^2}{1-{\cos }^2\theta }\left[\because {\sin }^2\theta +{\cos }^2\theta =1\right] \\ =\frac{{(1-\cos \theta )}^2}{(1-\cos \theta )(1+\cos \theta )}\left[\because a^2-b^2=(a-b)(a+b)\right] \\ =\frac{1-\cos \theta }{1+\cos \theta } \\ =\mathrm{RHS} \end{gathered}$$

Hence, proved.