# Answer on Trigonometry Question for sheetal

Question #17661

prove that cos10°- sin10° / cos 10°+ sin 10°= tan 35°

Expert's answer

(cos10°- sin10°) / ( cos 10°+ sin 10° )=( cos 10°+ sin10° )(cos10°- sin10°) / [( cos 10°+ sin 10° )( cos 10°+ sin 10° )]=

=(cos^2(10)-sin^2(10))/(sin^2(10)+2sin(10)cos(10)+cos^2(10))=cos(20)/( 1+sin(20) )=(1-sin(20))cos(20)/(1-sin^2(20))=

=(1-sin(20) )/cos(20)=1/cos(20) -tan(20) so we must showthat

1/cos(20) -tan(20)=tan(35)

1/cos(20) =tan(20)+tan(35)

but tan(x)+tan(y)=sin(x+y)/[cos(x)cos(y)]

tan(20)+tan(35)=sin(55)/[cos(20)*cos(35)]

so we must show that

1/cos(20) =sin(55)/[cos(20)*cos(35)]

sin(55)=cos(90-55)=cos(35)

and we have

1/cos(20)=sin(55)/[cos(20)*cos(35)]=sin(55)/[cos(20)*sin(55)]

1=1

=(cos^2(10)-sin^2(10))/(sin^2(10)+2sin(10)cos(10)+cos^2(10))=cos(20)/( 1+sin(20) )=(1-sin(20))cos(20)/(1-sin^2(20))=

=(1-sin(20) )/cos(20)=1/cos(20) -tan(20) so we must showthat

1/cos(20) -tan(20)=tan(35)

1/cos(20) =tan(20)+tan(35)

but tan(x)+tan(y)=sin(x+y)/[cos(x)cos(y)]

tan(20)+tan(35)=sin(55)/[cos(20)*cos(35)]

so we must show that

1/cos(20) =sin(55)/[cos(20)*cos(35)]

sin(55)=cos(90-55)=cos(35)

and we have

1/cos(20)=sin(55)/[cos(20)*cos(35)]=sin(55)/[cos(20)*sin(55)]

1=1

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