Question

if sin2x=n sin 2y,then the value of tan (x+y)/tan(x-y)​

Answer 1

please see the attachment

Answer 2

Given that,

We have,

$\sin2x=n\sin2y\\\\n=\dfrac{\sin (2x)}{\sin(2y)}$

To find,

Value of $\dfrac{\tan(x+y)}{\tan(x-y)}$

Solution,

Applying componendo and dividendo in above equation, we get :

$\dfrac{n+1}{n-1}=\dfrac{\sin2x+\sin2y}{\sin2x-\sin2y}$ .........(1)

Now using trigonometric formula :

$\sin A+\sin B=2\sin\dfrac{(A+B)}{2}{\cdot} \cos \dfrac{(A-B)}{2}\\\\\sin A-\sin B=2\cos\dfrac{(A+B)}{2}{\cdot} \sin\dfrac{(A-B)}{2}$

Using formula in equation (1) we get :

$\dfrac{n+1}{n-1}=\dfrac{2\sin ((2x+2y)/2){\cdot}\cos ((2x-2y)/2)}{2\cos((2x+2y)/2){\cdot}\sin((2x-2y)/2)}\\\\\dfrac{n+1}{n-1}=\dfrac{\sin (x+y){\cdot}\cos (x-y)}{\cos(x+y){\cdot}\sin(x-y)}\\\\\dfrac{n+1}{n-1}=\tan(x+y){\cdot}\cot(x-y)\\\\\dfrac{n+1}{n-1}=\dfrac{\tan(x+y)}{\tan(x-y)}$

So, the value of $\dfrac{\tan(x+y)}{\tan(x-y)}=\dfrac{n+1}{n-1}$

Learn more,

Trigonometry

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