Question

Prove that (cosec A − cot A)² = $\frac{(1-\cos A)}{(1+\cos A)}.$

Answer 1

(cosec A − cot A)² = $\frac{(1-\cos A)}{(1+\cos A)}.$

$$\begin{gathered} \text{LHS=}{(\cos \text{ec}A-\cot A)}^2 \\ \text{=}{(\frac{1}{\sin A}-\frac{\cos A}{\sin A})}^2 \\ \text{=}{\left(\frac{1-\cos A}{\sin A}\right)}^2 \\ \text{=}\frac{{(1-\cos A)}^2}{{\sin }^{2}A} \\ =\frac{{(1-\cos A)}^2}{1-{\cos }^{2}A}[\because {\sin }^2\theta +{\cos }^2\theta =1] \\ \text{=}\frac{(1-\cos A)(1-\cos A)}{(1-\cos A)(1+\cos A)} \\ \text{=}\frac{(1-\cos A)}{(1+\cos A)}\text{= RHS} \\ \text{Hence proved}\text{.} \end{gathered}$$