Question

Day 10,
Show that:
$\large \sf \dfrac{sin(x+y)}{sin(x-y)}=\dfrac{tan\:x+tan\:y}{tan\:x-tan\:y}$

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Answer 1

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

Consider RHS

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

can be rewritten as

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

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

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

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

Hence,

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

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Alternative Method

Consider LHS

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

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

On dividing numerator and denominator by cosx cosy,

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

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

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

Hence,

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

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Additional Information :-

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

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

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

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

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

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

Answer 2

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

Consider RHS

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

can be rewritten as

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

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

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

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

Hence,

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

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Alternative Method

Consider LHS

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

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

On dividing numerator and denominator by cosx cosy,

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

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

$\rm\green{ = \: \dfrac{tanx + tany}{tanx - tany}}$

Hence,

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

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Additional Information :-

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

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

$\green{\boxed{ \tt{ \: cos(x + y) = cosxcosy - sinysinx \: }}}$

$\pink{\boxed{ \tt{ \: cos(x - y) = cosxcosy + sinysinx \: }}}$

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

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

I hope it's helps you ❤️.

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