Math, asked by akayi929, 10 months ago

If sin(x+y)=cos(x+y) then prove tanx=1-tany/1+tany

Answers

Answered by DhanyaDA
3

Given

\sf sin(x+y)=cos(x+y)

To prove

\sf tanx=\dfrac{1-tany}{1+tany}

Explanation

 =  > sin(x + y) = cos(x + y)

 =  >  \dfrac{ sin(x + y)}{cos(x + y)} = 1

 =  > tan(x + y) = 1

Expanding using the formula

 \underline{ \sf \: tan(a + b) =  \dfrac{tana + tanb}{1 - tanatanb} }

then

 =  >  \dfrac{tanx + tany}{1 - tanxtany}  = 1

tanx + tany = 1 - tanxtany

 =  > tanx + tanxtany = 1 - tany

 =  > tanx(1 + tany) = 1 - tany

 =  > tanx =  \dfrac{1 - tany}{1 + tany}

 \boxed{ \sf \: tanx =  \dfrac{1 - tany}{1 + tany} }

Hence proved

Some more important formulas

 =  > sin(x + y) = sinxcosy + cosxsiny

 =  > cos(x + y) = cosxcosy - sinxsiny

 =  > tan( x - y) =  \dfrac{tanx - tany}{1 + tanxtany}

=>cot(x+y)=\dfrac{cotxcoty-1}{coty+cotx}

=>cot(x-y)=\dfrac{cotxcoty+1}{coty-cotx}

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