Question

Prove that: sin⁡(x+y) / sin(⁡x-y) = tan⁡ x + tan ⁡y / tan x -tan⁡ y

Answer 1

Answer:

Step-by-step explanation:

  From the properties of trigonometric ratios:

sin( A + B ) = sinAcosB + cosAsinB

sin( A - B ) = sinAcosB - cosAsinB

Here,

$\dfrac{sin(x+y)}{sin(x-y)}\\\\\\\implies \dfrac{sinx.cosy+siny.cosx}{sinx.cosy-siny.cosx}$

   

     Divide both denominator as well as numerator by cosx.cosy:

$\implies \dfrac{\dfrac{sinx.cosy+siny.cosx}{cosx.cosy}}{\dfrac{sinx.cosy+siny.cosx}{cosx.cosy}}\\\\\\\implies \dfrac{\dfrac{sinx.cosy}{cosx.cosy}+\dfrac{siny.cosx}{cosx.cosy}}{\dfrac{sinx.cosy}{cosx.cosy}+\dfrac{siny.cosx}{cosx.cosy}}\\\\\\\implies \dfrac{\dfrac{sinx}{cosy}+\dfrac{siny}{cosy}}{\dfrac{sinx}{cosy}-\dfrac{siny}{cosy}}\\\\\\\implies \dfrac{tanx + tany}{tanx - tany}$

  Hence proved

Answer 2

To ProvE :

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

SolutioN :

Taking LHS,

$\tt : \implies \dfrac{sin(x + y)}{sin(x - y)}$

Formula Use.

• Sin ( A + B ) = SinA.CosB + CosA.SinB.
• Sin( A - B ) = SinA.CosB - CosA.SinB.

$\tt : \implies \dfrac{sinx .cosy + cosx.siny}{sinx .cosy - cosx.siny}$

★ Dividing Cos x and Cos y both Numerator & Denominator.

$\tt : \implies \dfrac{ \Big(\frac{ sinx .cosy }{cosx.cosy}\Big)+ \Big( \frac{ cosx.siny}{cosx.cosy}\Big)}{\Big( \frac{ sinx .cosy}{cosx.cosy} \Big)- \Big( \frac{cosx.siny}{cosx.cosy}\Big)}$

$\tt : \implies \dfrac{ tanx + tany}{tanx - tany}$

ProveD.