Question

যদি tan x=a sin y ÷(1-a cos y) এবং tan y= b sin x÷(1-b cos x) হয়, তবে দেখাও যে, sin x÷sin y = a÷b​

Answer 1

$\bold{\pink{\underline{\underline{Given}}}}\longrightarrow \\ tanx = \dfrac{a \: siny}{1 - a \: cosy} \: \: \: \: and \\ \: tany = \dfrac{b \: sinx}{1 - b \: cosx}$

$\bold{\green{\underline{\underline{To \: prove}}}}\longrightarrow \\ \dfrac{sinx}{siny} = \dfrac{a}{b}$

$\bold{\blue{\underline{\underline{Concept \: used}}}}\longrightarrow \\ cot \alpha = \dfrac{1}{tan \alpha }$

$\bold{\red{\underline{\underline{Proof}}}}\longrightarrow \\ tanx = \dfrac{a \: siny}{1 - a \: cosy} \\ = > cotx = \dfrac{1 - a \: cosy}{a \: siny} \\ = > cotx = \dfrac{1}{a \: siny} - \dfrac{a \: cosy}{a \: siny} \\ = > cotx = \dfrac{1}{a \: sinx} - coty \\ = > cotx + coty = \dfrac{1}{a \: sinx} .....(1)$

$now$

$tany = \dfrac{b \: sinx}{1 - b \: cosx} \\ = > coty = \dfrac{1 - bcosx}{b \: sinx} \\ = > coty = \dfrac{1}{bsinx} - \dfrac{b \: cosx}{b \: sinx} \\ = > coty = \dfrac{1}{bsinx} - cotx \\ = > cotx + coty = \dfrac{1}{bsinx} .....(2)$

$by\:relation\:(1)\:and\:(2)\:we\:get$$\\\dfrac{1}{a\:siny}=\dfrac{1}{b\:sinx}\\=>b\:sinx=a\:siny\\=>\dfrac{sinx}{siny}=\dfrac{a}{b}$

Answer 2

Step-by-step explanation:

Given :

$\sf{tanx = \frac{a \: siny}{1 - a \: cosy} \: and \: tany = \frac{bsinx}{1 - bcosx}}$

To Prove :

$\sf{ \frac{ \sin(x) }{ \sin(y) }} = \frac{a}{b}$

Proof :

$\sf{ \tan(x )= \frac{a \sin(y) }{1 - a \cos(y) }} \\ \\ \\ \sf \implies \: \cot(x) = \frac{1 \: a \cos(y)}{a \sin(y) } \\ \\ \sf\implies \: \cot(x) = \frac{1}{a \sin(y) } - \frac{a \cos(y) }{a \sin(y) } \\ \\ \sf\implies \: \cot(x) = \frac{1}{a \sin(x) } - \cot(y) \\ \\ \sf \implies \: \cot(x )+ \cot(y) = \frac{1}{a \sin(x) } .......(1)$

$\sf{now} \\ \\ \sf \tan(y) = \frac{b \sin(x) }{1 - b \cos(x) } \\ \\ \sf \implies \: \cot(y) = \frac{1}{b \sin(x) } - \frac{b \cos(x) }{b \sin(x) } \\ \\ \sf \implies \: \cot(x) + \cot(y) = \frac{1}{b \sin(x) } .........(2) \\ \\ \sf{ \frac{1}{a \sin(y)}} \: = \frac{1}{b \sin(x) } \\ \\ \sf \implies \: \: \frac{ \sin(x) }{ \sin(y) } = \frac{a}{b}$