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`cot4x(sin5x+sin3x)= cotx(sin5x-sin3x)` |
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Answer» LHS `=cot4x(sin5x+sin3x)` `=(cos4x)/(sin4x)[2.sin(5x+3x)/(2) cos(5x-3x)/(2)]` `=(cos4x)/(sin4x).2sin4x.cosx.=2cos4x.cosx` `therefore cot4x(sin5x+sin3x)=2cos4xcosx`………………..(1) `=(cosx)/(sinx)(2.cos(5x+3x)/(2)sin(5x-3x)(2))` `=(cosx)/(sinx). 2cos4x.sinx=2cos4xcosx` `therefore cotx(sinx-sin3x)=2cos4xcosx`............(2) From equations, (1) and (2), `cot4x(sin5x+sin3x)=cotx(sin5x-sin3x)` LHS =RHS Hence proved. |
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