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Question
Show that : `sinAcosA - (sinAcos(90^circ - A)cosA)/sec(90^circ - A) - (cosAsin(90^circ - A)sinA)/(cosec(90^circ - A)) = 0`
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Solution
L.H.S. = `sinAcosA - (sinAsinAcosA)/(cosecA) - (cosAcosAsinA)/secA`
= sin A cos A – sin2 A cos A sin A – cos2 A sin A cos A
= sin A cos A – sin3 A cos A – cos3 A sin A
= sin A cos A [1 – sin2 A – cos2 A]
= sin A cos A [1 – (sin2 A + cos2 A)]
= sin A cos A (1 – 1)
= sin A cos A × 0
= 0 = R.H.S.
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