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Question: If 3 sin α = 5 sin β, then $\frac{tan \frac{\alpha+\beta}{2}}{tan \frac{\alpha-\beta}{2}}$ =...

If 3 sin α = 5 sin β, then tanα+β2tanα−β2\frac{tan \frac{\alpha+\beta}{2}}{tan \frac{\alpha-\beta}{2}} =

A

1

B

2

C

3

D

4

Answer

4

Explanation

Solution

Given the equation 3sin⁡α=5sin⁡β3 \sin \alpha = 5 \sin \beta.

We can rewrite this as: sin⁡αsin⁡β=53\frac{\sin \alpha}{\sin \beta} = \frac{5}{3}

Apply the Componendo and Dividendo rule: sin⁡α+sin⁡βsin⁡α−sin⁡β=5+35−3\frac{\sin \alpha + \sin \beta}{\sin \alpha - \sin \beta} = \frac{5+3}{5-3} sin⁡α+sin⁡βsin⁡α−sin⁡β=82=4\frac{\sin \alpha + \sin \beta}{\sin \alpha - \sin \beta} = \frac{8}{2} = 4

Now, use the sum-to-product and difference-to-product trigonometric identities: sin⁡A+sin⁡B=2sin⁡(A+B2)cos⁡(A−B2)\sin A + \sin B = 2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right) sin⁡A−sin⁡B=2cos⁡(A+B2)sin⁡(A−B2)\sin A - \sin B = 2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)

Substitute these into the equation: 2sin⁡(α+β2)cos⁡(α−β2)2cos⁡(α+β2)sin⁡(α−β2)=4\frac{2 \sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)}{2 \cos \left(\frac{\alpha+\beta}{2}\right) \sin \left(\frac{\alpha-\beta}{2}\right)} = 4

Simplify the expression: (sin⁡(α+β2)cos⁡(α+β2))⋅(cos⁡(α−β2)sin⁡(α−β2))=4\left( \frac{\sin \left(\frac{\alpha+\beta}{2}\right)}{\cos \left(\frac{\alpha+\beta}{2}\right)} \right) \cdot \left( \frac{\cos \left(\frac{\alpha-\beta}{2}\right)}{\sin \left(\frac{\alpha-\beta}{2}\right)} \right) = 4

This can be written in terms of tangent and cotangent: tan⁡(α+β2)⋅cot⁡(α−β2)=4\tan \left(\frac{\alpha+\beta}{2}\right) \cdot \cot \left(\frac{\alpha-\beta}{2}\right) = 4

Since cot⁡x=1tan⁡x\cot x = \frac{1}{\tan x}: tan⁡(α+β2)⋅1tan⁡(α−β2)=4\tan \left(\frac{\alpha+\beta}{2}\right) \cdot \frac{1}{\tan \left(\frac{\alpha-\beta}{2}\right)} = 4

Therefore, tan⁡(α+β2)tan⁡(α−β2)=4\frac{\tan \left(\frac{\alpha+\beta}{2}\right)}{\tan \left(\frac{\alpha-\beta}{2}\right)} = 4