Question

Prove that: $\frac{1-\mathrm{cos\theta }}{1+\mathrm{cos\theta }}=(\mathrm{cosec\theta }-\mathrm{cot\theta }{)}^2.$

Answer 1

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

LHS = $\frac{\left(1-\mathrm{cos\theta }\right)}{\left(1+\mathrm{cos\theta }\right)}$

Hence, LHS = RHS