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Question
Prove that:
`sin theta * cos theta {sin(pi/2 - theta) * "cosec" theta + cos (pi/2 - theta) * sec theta}` = 1
Solution
LHS = `sin theta * cos theta {sin(pi/2 - theta) * "cosec" theta + cos (pi/2 - theta) * sec theta}`
`= sin theta * cos theta {cos theta 1/(sin theta) + sin theta * 1/(cos theta)}`
`= sin theta * cos theta ((cos^2theta + sin^2theta)/(sin theta cos theta))`
= cos2θ + sin2θ = 1 = RHS ...[since sin2θ + cos2θ = 1]
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