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Question
Prove that:
`(sin(180^circ + "A")cos(90^circ - "A")tan(270^circ - "A"))/(sec(540^circ - "A") cos(360^circ + "A") "cosec"(270^circ + "A"))` = - sin A cos2 A.
Solution
LHS = `(sin(180^circ + "A")cos(90^circ - "A")tan(270^circ - "A"))/(sec(540^circ - "A") cos(360^circ + "A") "cosec"(270^circ + "A"))`
`= ((- sin "A")(sin "A")(cot "A"))/((- sec "A")(cos "A")(- sec "A"))` .....`[(sec (540° - "A")),(= sec (360° + 180° - "A")),(= sec (180° - "A")),(= (- sec "A"))]`
`= (- sin "A" sin "A" (cos "A")/(sin "A"))/(- 1/(cos "A")cos "A" - 1/(cos "A"))`
= - sin A × cos A × cos A
= - sin A cos2A = RHS
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