Question

if a=30 degree, verify that cos2a=1-tan square upon 1 tan square

Answer 1

so a=30°
$l.h.s = \cos(2 \times 30) \\ \: \: \: \: \: \: \: \: \: \: \: = \cos(60) = \frac{1}{2} \\ r.h.s = \frac{1 {tan}^{2}a }{1 - {tan}^{2} a} \\ = \frac{ {tan}^{2} 30 }{ 1 - {tan}^{2} 30} \\ = \frac{ { (\frac{1}{ \sqrt{3} } )}^{2} }{ 1 - {( \frac{1}{ \sqrt{3} } )}^{2} } \\ = \frac{ \frac{1}{3} }{ 1 - \frac{1}{3} } \\ = \frac{ \frac{1}{3} }{ \frac{2}{3} } \\ = \frac{1}{3} \times \frac{3}{2} \\ = \frac{1}{2} = r.h.s(proved)$
here is your answer...hope it helps... please please please mark it brainliest

Answer 2

Step-by-step explanation:

1-tansquarea/1+tansquarea