Question

Prove that:

`cos 80^circ/(sin 10^circ) + cos 59^circ "cosec"  31^circ = 2`

Answer 1

`"LHS"  = cos 80^circ/(sin 10^circ) + cos 59^circ "cosec"  31^circ `

`= (cos 80^circ)/cos(90^circ-10^circ) + sin (90^circ - 59^circ) "cosec"  31^circ`

`= 1 + "sin"31^circ`xx1/sin 31

= 1 + 1

= 2

= RHS