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CBSE Class 10 Answered

trigonometry
Asked by suchet_r | 05 Dec, 2009, 10:18: PM
answered-by-expert Expert Answer

(sinA - 2sin3 A)/(2cos3 A - cosA) =

sinA(1-2sin2A)/[cosA(2cos2A -1)] =

sinA(sin2A + cos2A  - 2sin2A)/[cosA(2cos2A -(sin2A + cos2A))] =

sinA(cos2A  - sin2A)/[cosA(cos2A -sin2A)] =

tanA

Regards,

Team,

TopperLearning.

Answered by | 06 Dec, 2009, 08:55: PM

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