Question

Prove this 1-tan^2x/1+tan^2x=cos^2x-sin^2x

Answer 1

Answer:

see explanation

Step-by-step explanation:

Using the identity

tan x = $\frac{sinx}{cosx}$

Consider the left side

$\frac{1-tan^2x}{1+tan^2x}$

= $\frac{1-\frac{sin^2x}{cos^2x} }{1+\frac{sin^2x}{cos^2x} }$ ( multiply numerator and denominator by cos²x to clear fractions

= $\frac{cos^2x-sin^2x}{cos^2x+sin^2x}$ ← ( cos²x + sin²x = 1 ]

= cos²x - sin²x

= right side , thus proven