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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