Math, asked by ketanpatil7600, 1 year ago

sin (x + y)/sin (x - y) = a+b/a-b then tan x/tan y =
1) b/a 2) a/b 3)1
4) 0​

Answers

Answered by jitekumar4201
0

Answer:

The option (b) is correct \dfrac{tanx}{tany} = \dfrac{a}{b}

Step-by-step explanation:

We have -

\dfrac{Sin (x+y)}{Sin(x-y)} = \dfrac{a+b}{a-b}

But sin ( x + y ) = sinx cosy + cosx siny

And sin ( x - y ) = sinx cosy - cosx siny

\dfrac{sinx \times cosy + cosx \times siny}{sinx \times cosy - cosx \times siny} = \dfrac{a+b}{a-b}

By Componando Dividendo Rule-

If \dfrac{a}{b} = \dfrac{c}{d}

Then \dfrac{a+b}{a-b} = \dfrac{c+d}{c-d}

\dfrac{sinx \times cosy + cosx \times siny + sinx \times cosy - cosx \times siny }{sinx \times cosy + cosx \times siny - sinx \times cosy +cosx \times siny}  = \dfrac{a+b+a-b}{a+b-a+b}

\dfrac{2sinx \times cosy}{2cosx \times siny} = \dfrac{2a}{2b}

\dfrac{sinx}{cosx}. \dfrac{cosy}{siny} = \dfrac{a}{b}

But \dfrac{sinx}{cosx} = tanx

And \dfrac{cosy}{siny} = \dfrac{1}{tany}

So \dfrac{tanx}{tany} = \dfrac{a}{b}

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