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Question
Prove that:
`(sin 70^circ)/(cos 20^circ) + ("cosec" 20^circ)/(sec 70^circ) - 2 cos 70^circ "cosec" 20^circ = 0`
Solution
LHS = `(sin 70^circ)/(cos 20^circ) + ("cosec" 20^circ)/(sec 70^circ) - 2 cos 70^circ "cosec" 20^circ`
`=("sin"70^circ)/(sin(90^circ - 20^circ)) + (sec(90^circ-20^circ))/sec 70^circ - 2 cos 70^circ sec(90^circ-20^circ)`
`= 1 + 1 - 2 xxcos 70^circxx1/cos 70^circ`
= 2 - 2
= 0
= RHS
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