P.T-(cosAcosB+sinAsinB)(cosAcosB-sinAsinB)=cos2^A-sin2^B

Asked by Prasanna Venkatesh Jayakumar | 27th Jul, 2010, 06:49: PM

Expert Answer:

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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