IF Tn=SINn A +COSnA prove that t3- t5/ t1= t5- t7 / t3

Asked by sonaisha | 24th Nov, 2008, 08:41: PM

Expert Answer:

LHS= sin3A+cos3A - sin5A+cos5A / (sinA+cosA)

=sin3A(1-sin2a) +cos3A (1-cos2A) / (sinA+cosA)

=sin3Acos2A+cos3Asin2A  / (sinA+cosA)

=cos2Asin2A (sinA+cosA) / (sinA+cosA)

=cos2Asin2A

similarly RHS = sin5A+cos5A - sin7A+cos7A / (sin3A+cos3A)

=sin5A(1-sin2a) +cos5A (1-cos2A) / (sin3A+cos3A)

=sin5Acos2A+cos5Asin2A  /(sin3A+cos3A)

=cos2Asin2A (sin3A+cos3A) / (sin3A+cos3A)

=cos2Asin2A

hence LHS=RHS

Answered by  | 30th Nov, 2008, 01:40: AM

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