Question

if cos 30°=√3÷2,prove that:-tan 15°=2-√3

Answer 1

Answer in the attachment.

_________________________________

Answer 2

Answer:

SOᒪᑌTIOᑎ

ᗯE ᕼᗩᐯE,

${tan}^{2} {15}^{0} = \frac{1 - {cos30}^{0} }{1 + {cos30}^{0} } \\ = \frac{1 - \frac{ \sqrt{3} }{2} }{1 + \frac{ \sqrt{3} }{2} } \\ = \frac{ \frac{2 - \sqrt{3} }{2} }{ \frac{2 + \sqrt{3} }{2} } \\ or \: {tan}^{2} {15}^{0} = \frac{2 - \sqrt{3} }{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } \\ or \: {tan}^{2} {15}^{0 } = \frac{ {(2 - \sqrt{3)} }^{2} }{4 - 3} \\ or \: {tan15}^{0} = \sqrt{ \frac{ {(2 - \sqrt{3)} }^{2} }{1} } \\ {tan15}^{0} = 2 - \sqrt{3 } \\ proved$