Question

If sin2a = k sin2b , prove that tan(a+b)/tan(a-b) = k+1/k-1 ​

Answer 1

To prove ->

tan (a+b)/tan(a-b) = k+1/k-1

Solution ->

As given that -> sin2a = k sin2b

we can write it as ->

$\frac{ \sin(2a) }{ \sin(2b) } = \frac{k}{1}$

Applying componendo-dividendo

$\frac{ \sin(2a) + \sin(2b) }{ \sin(2a) - \sin(2b) } = \frac{k + 1}{k - 1}$

As we know that

sinc + sind = sin (c+d/2).cos(c-d/2) and sinc -sind = cos(c+d/2).sin(c-d/2)

→ [sin(2a+2b/2)cos(2a-2b/2) ]/[cos(2a+2b/2)sin(2a-2b/2)] = k+1/k-1

→ sin(a+b)cos(a-b) / cos(a+b)sin(a-b) = k+1/k-1

→tan(a+b)cot(a-b) = k+1/k-1

We know that cot a = 1/tana ,so

→tan(a+b)×1/tan(a-b) = k+1/k-1

→tan(a+b)/tan(a-b). = k+1/k-1