if sinx=nsin(x+2y) prove that tan(x+y)=(1+n)/(1-n) tany

Asked by Rida Mukadam | 19th May, 2015, 07:15: PM

Expert Answer:

C o n s i d e r space t h e space g i v e n space e q u a t i o n colon sin x equals n sin open parentheses x plus 2 y close parentheses rightwards double arrow fraction numerator sin x over denominator sin open parentheses x plus 2 y close parentheses end fraction equals n A p p l y i n g space c o m p o n e n d o space d i v i d e n d o comma space w e space h a v e comma fraction numerator sin x plus sin open parentheses x plus 2 y close parentheses over denominator sin x minus sin open parentheses x plus 2 y close parentheses end fraction equals fraction numerator n plus 1 over denominator n minus 1 end fraction U sin g space t r i g o n o m e t r i c space i d e n t i t y sin C plus sin D equals 2 sin open parentheses fraction numerator C plus D over denominator 2 end fraction close parentheses cos open parentheses fraction numerator C minus D over denominator 2 end fraction close parentheses a n d sin C minus sin D equals 2 cos open parentheses fraction numerator C plus D over denominator 2 end fraction close parentheses sin open parentheses fraction numerator C minus D over denominator 2 end fraction close parentheses comma space i n space t h e space a b o v e space e q u a t i o n comma space w e space h a v e comma fraction numerator 2 sin begin display style fraction numerator x plus x plus 2 y over denominator 2 end fraction end style cos fraction numerator x minus open parentheses x plus 2 y close parentheses over denominator 2 end fraction over denominator 2 cos fraction numerator x plus x plus 2 y over denominator 2 end fraction sin fraction numerator x minus open parentheses x plus 2 y close parentheses over denominator 2 end fraction end fraction equals fraction numerator n plus 1 over denominator n minus 1 end fraction rightwards double arrow fraction numerator sin open parentheses begin display style x plus y end style close parentheses cos open parentheses negative y close parentheses over denominator 2 cos open parentheses x plus y close parentheses sin open parentheses negative y close parentheses end fraction equals fraction numerator n plus 1 over denominator n minus 1 end fraction rightwards double arrow negative fraction numerator sin open parentheses begin display style x plus y end style close parentheses cos y over denominator 2 cos open parentheses x plus y close parentheses sin y end fraction equals fraction numerator n plus 1 over denominator n minus 1 end fraction rightwards double arrow fraction numerator sin open parentheses begin display style x plus y end style close parentheses cos y over denominator 2 cos open parentheses x plus y close parentheses sin y end fraction equals negative open parentheses fraction numerator n plus 1 over denominator n minus 1 end fraction close parentheses rightwards double arrow fraction numerator sin open parentheses begin display style x plus y end style close parentheses cos y over denominator 2 cos open parentheses x plus y close parentheses sin y end fraction equals fraction numerator 1 plus n over denominator 1 minus n end fraction rightwards double arrow tan open parentheses x plus y close parentheses cross times c o t y equals fraction numerator 1 plus n over denominator 1 minus n end fraction rightwards double arrow tan open parentheses x plus y close parentheses equals open parentheses fraction numerator 1 plus n over denominator 1 minus n end fraction close parentheses tan y

Answered by Vimala Ramamurthy | 20th May, 2015, 09:04: AM

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