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| 1. |
sin0+cos0=p,sec0+cosec0=q.prove that q(p square-1)=2p (0=theta) |
| Answer» Given : cos{tex}\\theta{/tex} + sin{tex}\\theta{/tex}= p ....(i)\xa0sec{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)\xa0{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}\xa0{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}=\xa0{tex}2sin\\theta+2cos\\theta{/tex}= {tex}2(sin\\theta+cos\\theta){/tex}= {tex}2p{/tex} ......( using eq.i )= R.H.S.Hence Proved. | |