Question

If $\sf tanθ = \dfrac{x\: sinΦ}{1-x \:cosΦ}$ and $tanΦ=\dfrac{y \:sinθ}{1-y \:cos θ}$, find $\dfrac{x}{y}$​

Answer 1

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

Given that,

$\rm :\longmapsto\:tan\theta = \dfrac{xsin\phi }{1 - xcos\phi }$

and

$\rm :\longmapsto\:tan\phi = \dfrac{ysin\theta }{1 - ycos\theta }$

So, Consider

$\red{\rm :\longmapsto\:tan\theta = \dfrac{xsin\phi }{1 - xcos\phi }}$

$\red{\rm :\longmapsto\:tan\theta (1 - xcos\phi ) = xsin\theta }$

$\red{\rm :\longmapsto\:tan\theta - x \: tan\theta \: cos\phi = xsin\theta }$

$\red{\rm :\longmapsto\:tan\theta \: = x \: tan\theta \: cos\phi + xsin\theta }$

$\red{\rm :\longmapsto\:tan\theta \: = x \: (tan\theta \: cos\phi + sin\theta)}$

$\red{\rm :\longmapsto\:tan\theta \: = x \: \bigg(\dfrac{sin\theta }{cos\theta } \: cos\phi + sin\theta\bigg)}$

$\red{\rm :\longmapsto\:\dfrac{sin\theta }{cos\theta } \: = x \: \bigg(\dfrac{sin\theta \: cos\phi + sin\phi \: cos\theta }{cos\theta } \bigg)}$

We know

$\boxed{ \tt{ \: sinx \: cosy \: + \: siny \: cosx \: = \: sin(x + y) \: }}$

Thus,

$\red{\rm :\longmapsto\:sin\theta = x \: sin(\theta + \phi )}$

$\red{\rm \implies\:x = \dfrac{sin\theta }{sin(\theta + \phi )} - - - (1)}$

Now, Consider

$\purple{\rm :\longmapsto\:tan\phi = \dfrac{ysin\theta }{1 - ycos\theta } }$

$\purple{\rm :\longmapsto\:tan\phi (1 - ycos\theta ) = ysin\theta }$

$\purple{\rm :\longmapsto\:tan\phi - y \: tan\phi \: cos\theta = ysin\theta }$

$\purple{\rm :\longmapsto\:tan\phi = y \: tan\phi \: cos\theta + ysin\theta }$

$\purple{\rm :\longmapsto\:tan\phi = y( \: tan\phi \: cos\theta + sin\theta )}$

$\purple{\rm :\longmapsto\:tan\phi = y\bigg( \: \dfrac{sin\phi }{cos\phi } \: cos\theta + sin\theta \bigg)}$

$\purple{\rm :\longmapsto\:tan\phi = y\bigg( \: \dfrac{sin\phi cos\theta + sin\theta cos\phi }{cos\phi } \bigg)}$

$\purple{\rm :\longmapsto\:\dfrac{sin\phi }{cos\phi } = y\bigg( \: \dfrac{sin(\theta + \phi ) }{cos\phi } \bigg)}$

$\purple{\rm \implies\:y = \dfrac{sin\phi }{sin(\theta + \phi )} \: - - - (2)}$

So, On dividing equation (1) by (2), we get

$\blue{\rm \implies\:\boxed{ \bf{ \: \frac{x}{y} = \frac{sin\theta }{sin\phi } \: }}}$

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Additional Information

$\boxed{ \tt{ \: sinx \: cosy \: - \: siny \: cosx \: = \: sin(x - y) \: }}$

$\boxed{ \tt{ \: cosx \: cosy \: - \: siny \: sinx \: = \: cos(x + y) \: }}$

$\boxed{ \tt{ \: cosx \: cosy \: + \: siny \: sinx \: = \: cos(x - y) \: }}$

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

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

$\boxed{ \tt{ \: {sin}^{2}x - {sin}^{2}y = sin(x + y)sin(x - y) \: }}$

$\boxed{ \tt{ \: {cos}^{2}x - {sin}^{2}y = cos(x + y) \: cos(x - y) \: }}$