Question

Prove the following:

`sqrt(3)  "cosec"20^circ - sec20^circ` = 4

Answer 1

L.H.S. = `sqrt(3)  "cosec"20^circ - sec20^circ`

= `sqrt(3)/sin20^circ  - 1/cos20^circ`

= `2((sqrt(3)/2)/sin20^circ  - (1/2)/cos20^circ)`

= `2(sin60^circ/sin20^circ  - cos60^circ/cos20^circ)`

= `2((sin60^circ cos20^circ -  cos60^circ sin20^circ)/sin20^circ cos20^circ)`

= `(2sin(60^circ - 20^circ))/(1/2(2sin20^circ cos20^circ))`

= `(4sin40^circ)/(sin40^circ)`

= 4

= R.H.S.