Question

If tan alpha =x sin beta /1-x cos beta and tan beta =y sin alpha /1-y cos alpha, then proved that sin alpha /sin beta =x /y

Answer 1

Given :

tan α = $\dfrac{xsin\beta }{1-xcos\beta }$      and

tan β = $\dfrac{ysin\alpha }{1-ycos\alpha }$

To Prove :

$\dfrac{sin\alpha }{sin\beta }$  = $\dfrac{x}{y}$

Solution :

∵  tan α = $\dfrac{xsin\beta }{1-xcos\beta }$

Or,  $\dfrac{1}{tan\alpha }$ = $\dfrac{1}{\dfrac{xsin\beta }{1-ycos\beta } }$

Or,  cot α = $\dfrac{1-xcos\beta }{xsin\beta }$                                        ( ∵ tan α = $\dfrac{1}{cot\alpha }$ )

Or, cot α = $\dfrac{1}{xsin\beta }$ - $\dfrac{xcos\beta }{xsin\beta }$

Or, cot α = $\dfrac{1}{xsin\beta }$ - cot β

Or,  cotα + cotβ = $\dfrac{1}{xsin\beta }$              ...........1

Again

∵  tan β = $\dfrac{ysin\alpha }{1-ycos\alpha }$

Or,   $\dfrac{1}{tan\beta }$ = $\dfrac{1}{\dfrac{ysin\alpha }{1-ycos\alpha } }$

Or,  cot β = $\dfrac{1-ycos\alpha }{ysin\alpha }$                                        ( ∵ tan β = $\dfrac{1}{cot\beta }$ )

Or, cot β = $\dfrac{1}{ysin\alpha }$ - $\dfrac{ycos\alpha }{ysin\alpha }$

Or, cot β = $\dfrac{1}{ysin\alpha }$ - cot α

Or,  cotα + cotβ = $\dfrac{1}{ysin\alpha }$               ...........2

Now, from eq 1 and eq 2

$\dfrac{1}{xsin\beta }$    =  $\dfrac{1}{ysin\alpha }$

By cross multiplication

 $\dfrac{sin\alpha }{sin\beta }$  = $\dfrac{x}{y}$

Hence, $\dfrac{sin\alpha }{sin\beta }$  = $\dfrac{x}{y}$ is proved  . Answer