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`sin20^@(4+sec20^@)=` |
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Answer» `sin20^@(4+sec20^@)` `=sin20^@(4+1/cos20^@)` `=(4sin20^@cos20^@+sin20^@)/(cos20^@)` `=(2sin40^@+sin20^@)/(cos20^@)` `=(2sin(60-20)^@+sin20^@)/(cos20^@)` `=(2sin60^@cos20^@-2cos60^@sin20^@+sin20^@)/(cos20^@)` `=(2(sqrt3/2)cos20^@-2(1/2)sin20^@+sin20^@)/(cos20^@)` `=(sqrt3cos20^@-sin20^@+sin20^@)/(cos20^@)` `=sqrt3` So, `sin20^@(4+sec20^@) = sqrt3` |
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