Question

If $cosec\alpha +cot\alpha =p,$ prove that $cos\alpha =\frac{p^2-1}{p^2+1}.$

Answer 1

Given $\to cosec\alpha +cot\alpha =p$

To prove $\to cos\alpha =\frac{p^2-1}{p^2+1}$

$cosec\alpha +cot\alpha =\frac{1}{sin\alpha }+\frac{cos\alpha }{sin\alpha }=\frac{1+cos\alpha }{sin\alpha }$

So, $p=\frac{1+cos\alpha }{sin\alpha }$

RHS $\frac{p^2-1}{p^2+1}=\frac{(\frac{1+cos\alpha }{sin\alpha }{)}^2-1}{(\frac{1+cos\alpha }{sin\alpha }{)}^2+1}$

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

$=\frac{1+co{s}^2\alpha +2cos\alpha -1+co{s}^2\alpha }{1+co{s}^2\alpha +2cos\alpha +1-co{s}^2\alpha }$ $[si{n}^2\alpha +co{s}^2\alpha =1\Rightarrow si{n}^2\alpha -1-co{s}^2\alpha ]$

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

$=cos\alpha$

LHS = RHS

Hence proved.