Question

Prove that

$\frac{tanx - tany}{tanx + tany} = \frac{sin(x - y)}{sin(x + y)}$
Class ११ Trigonometry​

Answer 1

$\large\underline{\sf{Solution-}}$

Consider LHS

$\rm :\longmapsto\:\dfrac{tanx - tany}{tanx + tany}$

We know,

$\purple{\rm :\longmapsto\: \boxed{\tt{ \: tanx = \dfrac{sinx}{cosx}}}}$

So, using this identity, we get

$\rm \:  =  \: \dfrac{\dfrac{sinx}{cosx} - \dfrac{siny}{cosy} }{\dfrac{sinx}{cosx} + \dfrac{siny}{cosy} }$

$\rm \:  =  \: \dfrac{\dfrac{sinxcosy - sinycosx}{cosx \: cosy}}{ \: \: \: \dfrac{sinxcosy \: + \: sinycosx}{cosxcosy} \: \: \: }$

$\rm \:  =  \: \dfrac{sinxcosy - sinycosx}{sinxcosy + sinycosx}$

$\rm \:  =  \: \dfrac{sin(x - y)}{ \: \: sin(x + y) \: \: }$

Hence,

$\rm\implies \: \boxed{\tt{ \: \dfrac{tanx - tany}{ \: \: tanx + tany \: \: } = \frac{sin(x - y)}{ \: \: sin(x + y) \: \: } }}$

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More to know

$\boxed{\tt{ sin(x + y) = sinxcosy + sinycosx \: }}$

$\boxed{\tt{ sin(x - y) = sinxcosy - sinycosx \: }}$

$\boxed{\tt{ cos(x + y) = cosxcosy - sinxsiny}}$

$\boxed{\tt{ cos(x - y) = cosxcosy + sinxsiny}}$

$\boxed{\tt{ tan(x + y) = \frac{tanx + tany}{1 - tanxtany} }}$

$\boxed{\tt{ tan(x - y) = \frac{tanx - tany}{1 + tanxtany} }}$

Answer 2

$\large\underline{\sf{Solution-}}$

Consider LHS

$\rm :\longmapsto\:\dfrac{tanx - tany}{tanx + tany}$

We know,

$\purple{\rm :\longmapsto\: \boxed{\tt{ \: tanx = \dfrac{sinx}{cosx}}}}$

So, using this identity, we get

$\rm \: = \: \dfrac{\dfrac{sinx}{cosx} - \dfrac{siny}{cosy} }{\dfrac{sinx}{cosx} + \dfrac{siny}{cosy} }$

$\rm \: = \: \dfrac{\dfrac{sinxcosy - sinycosx}{cosx \: cosy}}{ \: \: \: \dfrac{sinxcosy \: + \: sinycosx}{cosxcosy} \: \: \: }$

$\rm \: = \: \dfrac{sinxcosy - sinycosx}{sinxcosy + sinycosx }$

$\rm \: = \: \dfrac{sin(x - y)}{ \: \: sin(x + y) \: \: }$

Hence,

$\rm\implies \: \boxed{\tt{ \: \dfrac{tanx - tany}{ \: \: tanx + tany \: \: } = \frac{sin(x - y)}{ \: \: sin(x + y) \: \: } }}$

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More to know

$\boxed{\tt{ sin(x + y) = sinxcosy + sinycosx \: }}$

$\boxed{\tt{ sin(x - y) = sinxcosy - sinycosx \: }}$

$\boxed{\tt{ cos(x + y) = cosxcosy - sinxsiny}}$

$\boxed{\tt{ cos(x - y) = cosxcosy + sinxsiny}}$

$\boxed{\tt{ tan(x + y) = \frac{tanx + tany}{1 - tanxtany} }}$

$\boxed{\tt{ tan(x - y) = \frac{tanx - tany}{1 + tanxtany} }}$

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