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Prove the following trigonometric identity
Asked by | 16 Apr, 2010, 07:08: PM Expert Answer

cosecx - sinx = a3

1/sinx - sinx = a3

(1 - sin2x)/sinx = a3

cos2x/sinx = a3

(cos2x/sinx)2/3 = a2

Similarly,

(sin2x/cosx)2/3 = b2

Now, a2b2(a2+b2) =

[(cos2x/sinx)(sin2x/cosx)]2/3{(cos2x/sinx)2/3 + (sin2x/cosx)2/3} =

[cosx sinx]2/3{[(cos3x)2/3 + (sin3x)2/3 ]/(sinxcosx)2/3} =

[cosx sinx]2/3{[cos2x + sin2x ]/(sinxcosx)2/3} =

[cosx sinx]2/3{1/(sinxcosx)2/3} = 1.

Regards,

Team,

TopperLearning.

Answered by | 16 Apr, 2010, 08:58: PM

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