Question

Solve this question as soon as possible.​

Answer 1

Step-by-step explanation:

Given

$\sin( \frac{ \alpha }{2} ) = \sqrt{ \frac{x - 1}{2x} }$

Then

$\cos( \frac{ \alpha }{2} ) = \sqrt{ \frac{x + 1}{2x} }$

$we \: know \: that \: \\ \sin( \alpha ) = 2 \sin( \frac{ \alpha }{2} ) \cos( \frac{ \alpha }{2} )$

$\sin( \alpha ) = 2 \times \sqrt{ \frac{x - 1}{2x} } \times \sqrt{ \frac{x + 1}{2x} }$

$\sin( \alpha ) = 2 \times \sqrt{ \frac{(x - 1)(x + 1)}{2x \times 2x} }$

$\sin( \alpha ) = \sqrt{ \frac{ {x}^{2} - 1}{ {x}^{2} } }$

$\tan( \alpha ) = \sqrt{ \frac{ {x}^{2} - 1}{ 1}}$