sin0+cos0=p,sec0+cosec0=q.prove that q(p square-1)=2p (0=theta)

Given : cos{tex}\theta{/tex} + sin{tex}\theta{/tex}= p ....(i)

 sec{tex}\theta{/tex} + cosec{tex}\theta{/tex} = q....(ii)
L.H.S = q(p²- 1)
= (sec{tex}\theta{/tex} + cosec{tex}\theta{/tex}) [(cos{tex}\theta{/tex} + sin{tex}\theta{/tex})² - 1]
= (sec{tex}\theta{/tex} + cosec{tex}\theta{/tex}) (cos²{tex}\theta{/tex} + sin²{tex}\theta{/tex} + 2 sin{tex}\theta{/tex} cos{tex}\theta{/tex} - 1) {tex}[\because (a+b)^2=a^2+b^2+2ab]{/tex}
{tex}= \left( \frac { 1 } { \cos \theta } + \frac { 1 } { \sin \theta } \right) ( 2 \sin \theta \cos \theta ){/tex} {tex}[\because sin^2\theta+cos^2\theta=1]{/tex}
{tex}= 2sin\theta cos\theta. \frac{1}{cos\theta}+2sin\theta cos\theta.\frac{1}{sin\theta}{/tex}
= {tex}2sin\theta+2cos\theta{/tex}
= {tex}2(sin\theta+cos\theta){/tex}
= {tex}2p{/tex}   ......( using eq.i )
= R.H.S.
Hence Proved.