P.T-(cosAcosB+sinAsinB)(cosAcosB-sinAsinB)=cos2^A-sin2^B
Asked by Prasanna Venkatesh Jayakumar | 27th Jul, 2010, 06:49: PM
Dear Student,
Here the L.H.S. is
(cosAcosB + sinAsinB)(cosAcosB - sinAsinB)
= (cosAcosB)2 - (sinAsinB)2
= (cos2A)(cos2B) - (sin2A)(sin2B)
= cos2A(1 – sin2B) – (1 – cos2A)sin2B .........{Because sin2A + cos2A = 1}
= cos2A – cos2A sin2B – sin2B + cos2Asin2B
= cos2A – sin2B
= R.H.S
Hence the result is proved
Regards,
Team Topper Learning
Dear Student,
Here the L.H.S. is
(cosAcosB + sinAsinB)(cosAcosB - sinAsinB)
= (cosAcosB)2 - (sinAsinB)2
= (cos2A)(cos2B) - (sin2A)(sin2B)
= cos2A(1 – sin2B) – (1 – cos2A)sin2B .........{Because sin2A + cos2A = 1}
= cos2A – cos2A sin2B – sin2B + cos2Asin2B
= cos2A – sin2B
= R.H.S
Hence the result is proved
Regards,
Team Topper Learning
Answered by | 27th Jul, 2010, 08:31: PM
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