Question

Prove that cos2α = 1-tan²α / 1+tan²α​

Answer 1

Answer:

see the explanation

Step-by-step explanation:

we have

$cos(2\alpha)=\frac{1-tan^2(\alpha)}{1+tan^2(\alpha)}$

Prove

Remember that

$cos(2\alpha)=cos^2(\alpha)-sin^2(\alpha)$

$tan(\alpha)=\frac{sin(\alpha)}{cos(\alpha)}$

substitute in the expression above

$cos^2(\alpha)-sin^2(\alpha)=\frac{1-(\frac{sin(\alpha)}{cos(\alpha)})^2}{1+(\frac{sin(\alpha)}{cos(\alpha)})^2}$    

$cos^2(\alpha)-sin^2(\alpha)=\frac{(\frac{cos^2(\alpha)-sin^2(\alpha)}{cos^2(\alpha)}}{(\frac{cos^2(\alpha)+sin^2(\alpha)}{cos^2(\alpha)}}}$

Remember that

$cos^2(\alpha)+sin^2(\alpha)=1$

Simplify

$cos^2(\alpha)-sin^2(\alpha)=cos^2(\alpha)-sin^2(\alpha)$ ---> is proved