Question

Sin a+b /sin a-b = x+y/x-y then show that x tanb=y tan a

Answer 1

$\mathsf{Given :\;\dfrac{sin(a + b)}{sin(a - b)} = \dfrac{x + y}{x - y}}$

$\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{sin(A + B) = sinA.cosB + cosA.sinB}}}$

$\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{sin(A - B) = sinA.cosB - cosA.sinB}}}$

$\mathsf{\implies \dfrac{sina.cosb + cosa.sinb}{sina.cosb - cosa.sinb} = \dfrac{x + y}{x - y}}$

Applying Componendo and Dividendo Theorem on both sides :

$\mathsf{\implies \dfrac{(sina.cosb + cosa.sinb) + (sina.cosb - cosa.sinb)}{(sina.cosb + cosa.sinb) - (sina.cosb - cosa.sinb)} = \dfrac{(x + y) + (x - y)}{(x + y) - (x - y)}}$

$\mathsf{\implies \dfrac{2sina.cosb}{(sina.cosb + cosa.sinb - sina.cosb + cosa.sinb} = \dfrac{2x}{(x + y - x + y)}}$

$\mathsf{\implies \dfrac{2sina.cosb}{2cosa.sinb} = \dfrac{2x}{2y}}$

$\mathsf{\implies \dfrac{sina.cosb}{cosa.sinb} = \dfrac{x}{y}}$

$\mathsf{\implies \bigg(\dfrac{sina}{cosa}\bigg)\bigg(\dfrac{cosb}{sinb}\bigg)= \dfrac{x}{y}}$

$\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{tan\theta = \dfrac{sin\theta}{cos\theta}}}}$

$\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{cot\theta = \dfrac{cos\theta}{sin\theta}}}}$

$\mathsf{\implies tana.cotb = \dfrac{x}{y}}$

$\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{cot\theta = \dfrac{1}{tan\theta}}}}$

$\mathsf{\implies \dfrac{tana}{tanb} = \dfrac{x}{y}}$

$\implies \large\boxed{\mathsf{x.tanb = y.tana}}$

Answer 2

Step-by-step explanation:

$\mathsf{Given :\;\dfrac{sin(a + b)}{sin(a - b)} = \dfrac{x + y}{x - y}}$

$\mathsf{\implies \dfrac{sina.cosb + cosa.sinb}{sina.cosb - cosa.sinb} = \dfrac{x + y}{x - y}}$

Applying Componendo and Dividendo Theorem on both sides :

$\mathsf{\implies \dfrac{(sina.cosb + cosa.sinb) + (sina.cosb - cosa.sinb)}{(sina.cosb + cosa.sinb) - (sina.cosb - cosa.sinb)} = \dfrac{(x + y) + (x - y)}{(x + y) - (x - y)}}$

$\mathsf{\implies \dfrac{2sina.cosb}{(sina.cosb + cosa.sinb - sina.cosb + cosa.sinb} = \dfrac{2x}{(x + y - x + y)}}$

$\mathsf{\implies \dfrac{2sina.cosb}{2cosa.sinb} = \dfrac{2x}{2y}}$

$\mathsf{\implies \dfrac{sina.cosb}{cosa.sinb} = \dfrac{x}{y}}$

$\mathsf{\implies \bigg(\dfrac{sina}{cosa}\bigg)\bigg(\dfrac{cosb}{sinb}\bigg)= \dfrac{x}{y}}$

$\mathsf{\implies tana.cotb = \dfrac{x}{y}}$

$\mathsf{\implies \dfrac{tana}{tanb} = \dfrac{x}{y}}$

$\implies \large\boxed{\mathsf{x.tanb = y.tana}}$