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| 1. |
a cosΦ +b sinΦ=ma sin Φ -b cosΦ=n |
| Answer» Given,R.H.S = m2 +n2= (a cos{tex}\\theta{/tex}\xa0+ b sin{tex}\\theta{/tex})2 + (a sin{tex}\\theta{/tex}\xa0- b cos{tex}\\theta{/tex})2 {tex}\\left[\\sin ce,\\;m\\;=\\;a\\;\\cos\\theta\\;+\\;b\\;\\sin\\theta\\;and\\;n\\;=\\;a\\;\\sin\\theta\\;-\\;b\\;\\cos\\theta\\right]{/tex}\xa0= (a2cos2{tex}\\theta{/tex}\xa0+ b2sin2{tex}\\theta{/tex}\xa0+\xa02ab cos{tex}\\theta{/tex}\xa0sin{tex}\\theta{/tex}) + (a2sin2{tex}\\theta{/tex}\xa0+ b2cos2{tex}\\theta{/tex}\xa0- 2ab sin{tex}\\theta{/tex}\xa0cos{tex}\\theta{/tex}) {tex}\\left[\\because\\left(a\\pm b\\right)^2=a^2+b^2\\pm2ab\\right]{/tex}= a2 (cos2{tex}\\theta{/tex}\xa0+ sin2{tex}\\theta{/tex}) + b2 (sin2{tex}\\theta{/tex}\xa0+ cos2{tex}\\theta{/tex}) = a2 + b2 = L.H.S {tex}\\left[\\because\\sin^2\\theta+\\cos^2\\theta=1\\right]{/tex}therefore, {tex}m^2\\;+n^2=a^2+b^2{/tex}Hence proved. | |