Trigonometry

Asked by  | 12th Feb, 2010, 09:39: AM

Expert Answer:

(a2+b2)3 = (a2)3 + (b2)3 + 3a2b2(a2+b2)

3a2b2(a2+b2) = (a2+b2)3 - (a2)3 + (b2)3

Now,

cosecA - sinA = a3

cos2A/sinA = a3

Similarly,

b3 = sin2A/cosA

a2 = (cos2A/sinA)2/3 =  cos4/3A/sin2/3A

b2 = (sin2A/cosA)2/3 = sin4/3A/cos2/3A

Hence,

3a2b2(a2+b2) = (cos4/3A/sin2/3A + sin4/3A/cos2/3A)3 - (cos4A/sin2A) - (sin4A/cos2A)

= ((cos6/3A + sin6/3A)/sin2/3Acos2/3A)3 - ((cos6A + sin6A)/sin2Acos2A)

= ((cos2A + sin2A)/sin2/3Acos2/3A)3 - ((cos6A + sin6A)/sin2Acos2A)

= (1/sin2/3Acos2/3A)3 - ((cos6A + sin6A)/sin2Acos2A)

= (1/sin2Acos2A) - ((cos6A + sin6A)/sin2Acos2A)

= (1 - (cos6A + sin6A))/(sin2Acos2A)

= (cosec2Asec2A - cosec2Acos4A - sin4Asec2A)

a2b2(a2+b2) = (cosec2Asec2A - cosec2Acos4A - sin4Asec2A)/3

a4b2 + a2b4 = (cosec2Asec2A - cosec2Acos4A - sin4Asec2A)/3

Regards,

Team,

TopperLearning.

 

 

Answered by  | 12th Feb, 2010, 08:23: PM

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