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{ (1-sinAcosA)/cosA(secA-cosecA) } . [ {(sinA-cosA) (sinA+cosA)}/ sin cube A+ cos cube A] =sin A

asked Aug 23, 2014 in TRIGONOMETRY by anonymous

1 Answer

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Left hand side identityimage

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= Right hand side identity.

answered Aug 23, 2014 by david Expert

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