Question

Prove the following identities , where the angles involved are acute angles for which the expressions are defined .
(i) ${(cosec\theta -cot\theta )}^2=\frac{1-cos\theta }{1+cos\theta }$
(ii) $\frac{cosA}{1+sinA}+\frac{1+sinA}{cosA}=2secA$

Answer 1

$\left(i\right)(cosec\theta -cot\theta {)}^2=\frac{1-cos\theta }{1+cos\theta }$

LHS

$(cosec\theta -cot\theta {)}^2=(\frac{1}{sin\theta }-\frac{cos\theta }{sin\theta }{)}^2$

$=\frac{(1-cos\theta {)}^2}{(sin\theta {)}^2}=\frac{1+co{s}^2\theta -2cos\theta }{si{n}^2\theta }$

$=\frac{1-co{s}^2\theta +2co{s}^2\theta -2cos\theta }{si{n}^2\theta }$

$=\frac{(1-co{s}^2\theta )}{(1-co{s}^2\theta )}+\frac{2cos\theta (cos\theta -1)}{(1-co{s}^2\theta )}$

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

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

$\therefore LHS=RHS$

(ii)$\frac{cosA}{(1+sinA)}+\frac{1+sinA}{cosA}=2secA.$

LHS

$\Rightarrow \frac{co{s}^{2}A+(1+sinA{)}^2}{\left(cosA\right)(1+sinA)}\Rightarrow \frac{co{s}^{2}A+si{n}^{2}A+2sinA+1}{\left(cosA\right)(1+sinA)}$

$\Rightarrow \frac{1+2sinA+1}{\left(cosA\right)(1+sinA)}\Rightarrow \frac{2(1+sinA)}{cosA(1+sinA)}$

$\Rightarrow \frac{2}{cosA}=2secA=RHS$

$\therefore LHS=RHS$