Question

prove that =cosA=(1-tan^2A/2)/(1+tan^2A/2)

Answer 1

Hey there !!

We have ,

$\frac{ {1 - \tan {}^{2} ( a) }}{2} \: \div \frac{ 1 + \: \tan^{2} \:a}{2} \\ \\\frac{ {1 - \tan \: ^{2} a }}{2} \: \: \times \frac{2}{ \: 1 + { \tan}^{2}a \: } \\ \\ \: \frac{1 - { \tan}^{2}a }{ {1 + \tan}^{2}a }$
Putting → Tan ²A = Sin²A / cos²A

so , we get

{ 1 - Sin²A / cos²A } / { 1 + Sin²A / cos²A }

{ Cos²A - sin²A / Cos²a } / cos² A+ sin²A / cos²A

( Cos² A and Cos²A will be cancelled out )

Now ,

Cos² A- sin²A / ( ∵ sin²A + cos²A = 1 )

so ,

cos 2A

Dividing by 2 we get
cosA

⋆ Hope this would help you ⋆

Answer 2

Hey friends ....

As you know that cos2A=cos²A-sin²A

and....we can also write that ..

cosA=cos²A/2-sin²A/2/1 ...If we put 1 on denominator then there is no change .

So

CosA=cos²A/2-sin²A/2/cos²A/2+sin²A/2 ,

{as you know that cos²A+cos²A/2=1 similarly here -Ais also )

CosA=cos²A-sin²A/cos²A/2÷cos²A/2+sin²A/2/cos²A/2

dividing by cos²A/2 on numerator and denominator)

cosA=1-tan²A/1+tan²A/2 prooved here ..

hope it helps you

@Rajukumar