Question

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

Answer 1

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

given $\Rightarrow cosec\theta +tan\theta =p$

$\Rightarrow$ square both the side of the given equation

$\therefore (coseco\theta +cot\theta {)}^2=p^2$

$\therefore cose{c}^2\theta +co{t}^2\theta +2cosec.cot\theta =p^2$

$\therefore \frac{cose{c}^2\theta +co{t}^2\theta +2cosec\theta .cot\theta +1}{cose{c}^2\theta +co{t}^2\theta +2cosec\theta .cot\theta -1}=\frac{p^2+1}{p^2-1}$ ($\therefore$ using compound o & dividend rule)

$\therefore \frac{2cosec\theta +2cosec\theta .cot\theta }{2co{t}^2\theta +2cosec\theta .cot\theta }=\frac{p^2+1}{p^2-1}$ $(\because 1+co{t}^2\theta =cose{c}^2\theta )$

$\therefore \frac{2csec\theta (cose\theta +cot\theta )}{2cot\theta (cot\theta +cosec\theta )}=\frac{p^2+1}{p^2-1}$

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

$\therefore \frac{1}{sin\theta }.\frac{sin\theta }{coos\theta }=\frac{p^2+1}{p^2-1}\to \frac{1}{cos\theta }=\frac{p^2+1}{p^2-1}$

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

Hence Proved