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Question
Prove the following:
`sqrt(3) "cosec"20^circ - sec20^circ` = 4
Solution
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.
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