Question

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

Answer 1

`(1 - \tan^2(x))/(1 + \tan^2(x))\\= ((\cos^2(x))/(\cos^2(x)) - (\sin^2(x))/(\cos^2(x)))/((\cos^2(x))/(\cos^2(x)) + (\sin^2(x))/(\cos^2(x)))\\= ((\cos^2(x) - \sin^2(x))/(\cos^2(x)))/((\cos^2(x) + \sin^2(x))/(\cos^2(x)))\\= (\cos^2(x) - \sin^2(x))/(\cos^2(x)) \cdot (\cos^2(x))/(\cos^2(x) + \sin^2(x))\\= ((\cos^2(x) - \sin^2(x)) \cdot \cos^2(x))/(\cos^2(x) \cdot (\cos^2(x) + \sin^2(x)))\\= (\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

see explanation

Explanation:

Using the trigonometric identities

tan x =
`(sinx)/(cosx)` , sin²x + cos²x = 1

Consider the left side

`(1-tan^2x)/(1+tan^2x)`

=
`(1-(sin^2x)/(cos^2x) )/(1+(sin^2x)/(cos^2x) )` ← multiply numerator/ denominator of the whole fraction by cos²x

=
`(cos^2x-sin^2x)/(cos^2x+sin^2x)`

=
`(cos^2x-sin^2x)/(1)`

= cos²x - sin²x

= cos²x - (1 - cos²x)

= cos²x - 1 + cos²x

= 2cos²x - 1

= right side , thus proven