Question

Prove that:

`(sin 70^circ)/(cos 20^circ) + ("cosec" 20^circ)/(sec 70^circ) - 2  cos 70^circ "cosec"  20^circ = 0`

Answer 1

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