Question

If $\cos ec\theta +\cot \theta =p$, then prove that $\cos \theta =\frac{P^2-1}{p^2+1}$

Answer 1

We know that

$cose{c}^2-co{t}^2=1$

$(cosec-cot)\times (cosec+cot)=1$

$cosec-cot=\left(1/p\right)$

$cosec+cot=p$

$2cosec=p+1/p$

$2cosec=(p^2+1)/p$

$cosec=(p^2+1)/2p$

$sin=2p/(p^2+1)$

$si{n}^2=4p^2/\left(\right(p^2+1{)}^2)$

$1-si{n}^2=co{s}^2=\left(\right(p^2-1{)}^2\left)/\right((p^2+1{)}^2)$

$cos=(p^2-1)/(p^2+1)$

Hence Proved