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Show that `sqrt(3)cosec 20^@ - sec 20^@= 4` |
Answer» `sqrt3 cosec 90^@ - sec 20^@ = 4` LHS `sqrt3 cosec 90^@ - sec 20^@` `= sqrt3 1/(sin 20^@) -1/(cos 20^@) = ( sqrt3 cos 90^@ - sin 20^@)/(sin20^@* cos20^@)` `= (sqrt3 cos 20^@ - sin 20^@)/(sin 20^@ * cos 20^@)` `= (sqrt3/2cos 20^@ - 1/2 sin 20^@)/(1/2 sin 20^@ cos20^@)` `= (sin 60^@cos20^@ - cos 60^@sin20^@)/(1/2 sin 20^@cos20^@)` `= 2 xx 2 xx (sin (60^@-20^@))/(2 sin20^@cos90^@)` `= 4 xx (sin 40^@)/(sin 40^@)` `= 4 =`RHS Hence proved |
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