Question

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

Answer 1

Given, $cosec\theta +cot\theta =p$

$\Rightarrow \frac{1}{sin\theta }+\frac{cos\theta }{sin\theta }=p[\because cosec\theta =\frac{1}{sin\theta }andcot\theta =\frac{cos\theta }{sin\theta }]$

$\Rightarrow \frac{1+cos\theta }{sin\theta }=\frac{p}{1}$

$\Rightarrow \frac{(1+cos\theta {)}^2}{si{n}^2\theta }=\frac{p^2}{1}$ [take square on both sides]

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

Using componendo and dividendo rule, we get;

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

$\Rightarrow \frac{1+co{s}^2\theta +2cos\theta -(1-co{s}^2\theta )}{1+2cos\theta +(co{s}^2\theta +si{n}^2\theta )}=\frac{p^2-1}{p^2+1}$ $[\because si{n}^2\theta +co{s}^2\theta =1]$

$\Rightarrow \frac{2co{s}^2\theta +2cos\theta }{2+2cos\theta }=\frac{p^2-1}{p^2+1}$

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

$\therefore cos\theta =\frac{p^2-1}{p^2+1}$