If sin-sin-a-ba-b then prove that a tan-b tan-

We have, sin(α +β )sin (α β )=a+ba b Applying componendo and dividendo sin (α +β )+sin (α β )sin (α +β ) sin (α β )=a+b+a ba+b (a b) sin C+ sin ...

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Trigonometric Ratios of Compound Angles

If sinα+βsi...

Question

If $\frac{sin (α + β)}{sin (α - β)} = \frac{a + b}{a - b}$ then prove that $a tan β = b tan α$

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\sin α + \sin β = a and \cos α + \cos β = b

\sin (α + β) = \frac{2 a b}{a^{2} + b^{2}}

\cos (α - β) = \frac{a^{2} + b^{2} - 2}{2}

s i n α + s i n β = a

c o s α + c o s β = b

sin (α + β) = \frac{2 a b}{a^{2} + b^{2}}

\frac{sin (α + β)}{sin (α - β)} = \frac{a + b}{a - b}

α \neq β, a \neq b, b \neq 0

\frac{tan α}{tan β}

0 < α, β, γ < \frac{π}{2}

sin α + sin β + sin γ > sin (α + β + γ)

sin α + sin β = a, cos α + cos β = b

sin (α + β)

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