Solveeit Logo
Login

Question

Question: \(\tan\frac{A}{2}\)is equal to...

tan⁡A2\tan\frac{A}{2}is equal to

A

±1−sin⁡A1+sin⁡A\pm \sqrt{\frac{1 - \sin A}{1 + \sin A}}

B

±1+sin⁡A1−sin⁡A\pm \sqrt{\frac{1 + \sin A}{1 - \sin A}}

C

±1−cos⁡A1+cos⁡A\pm \sqrt{\frac{1 - \cos A}{1 + \cos A}}

D

±1+cos⁡A1−cos⁡A\pm \sqrt{\frac{1 + \cos A}{1 - \cos A}}

Answer

±1−cos⁡A1+cos⁡A\pm \sqrt{\frac{1 - \cos A}{1 + \cos A}}

Explanation

Solution

tan⁡(A2)=sin⁡(A/2)cos⁡(A/2)=±(1−cos⁡A)/2(1+cos⁡A)/2=±1−cos⁡A1+cos⁡A\tan\left( \frac{A}{2} \right) = \frac{\sin(A/2)}{\cos(A/2)} = \pm \sqrt{\frac{(1 - \cos A)/2}{(1 + \cos A)/2}} = \pm \sqrt{\frac{1 - \cos A}{1 + \cos A}}.