Question

Prove that:

`sin theta * cos theta {sin(pi/2 - theta) * "cosec" theta + cos (pi/2 - theta) * sec theta}` = 1

Answer 1

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]