Question

Prove that 1-tan^2x/1+tan^2x=2cos^2x-1​

Answer 1

$\frac{1 - \tan^2(x)}{1 + \tan^2(x)}\\= \frac{\frac{\cos^2(x)}{\cos^2(x)} - \frac{\sin^2(x)}{\cos^2(x)}}{\frac{\cos^2(x)}{\cos^2(x)} + \frac{\sin^2(x)}{\cos^2(x)}}\\= \frac{\frac{\cos^2(x) - \sin^2(x)}{\cos^2(x)}}{\frac{\cos^2(x) + \sin^2(x)}{\cos^2(x)}}\\= \frac{\cos^2(x) - \sin^2(x)}{\cos^2(x)} \cdot \frac{\cos^2(x)}{\cos^2(x) + \sin^2(x)}\\= \frac{(\cos^2(x) - \sin^2(x)) \cdot \cos^2(x)}{\cos^2(x) \cdot (\cos^2(x) + \sin^2(x))}\\= \frac{\cos^2(x) - \sin^2(x)}{\cos^2(x) + \sin^2(x)}$

$= \cos^2(x) - \sin^2(x)\\= \cos^2(x) - (1 - \cos^2(x))\\= \cos^2(x) - 1 + \cos^2(x)\\= 2\cos^2(x) - 1$

Answer 2

Answer:

see explanation

Step-by-step explanation:

Using the trigonometric identities

tan x = $\frac{sinx}{cosx}$ , sin²x + cos²x = 1

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/ denominator of the whole fraction by cos²x

= $\frac{cos^2x-sin^2x}{cos^2x+sin^2x}$

= $\frac{cos^2x-sin^2x}{1}$

= cos²x - sin²x

= cos²x - (1 - cos²x)

= cos²x - 1 + cos²x

= 2cos²x - 1

= right side , thus proven