Question

How do you verify `cos^2x + tan^2x - 1 / cos^2x + sin^2x = 0`?

Answer 1

Please see below.

Explanation:

We know that,

`color(blue)((1)1/costheta=sectheta`

`color(red)((2)sin^2theta+cos^2theta=1`

`color(red)((3)sec^2theta-tan^2theta=1`

We have to verify :

`cos^2x+tan^2x-color(blue)(1/cos^2x)+sin^2x=0`

We take left hand side :

`LHS=cos^2x+tan^2x-color(blue)(1/cos^2x)+sin^2x...tocolor(blue)(Apply(1)`

`LHS=cos^2x+tan^2x-color(blue)(sec^2x)+sin^2x`

`LHS=cos^2x+sin^2x-sec^2x+tan^2x`

`LHS={color(red)(cos^2x+sin^2x)}-{color(red)(sec^2x-tan^2x)}`

`LHS=1-1...tocolor(red)(Apply(2) and (3)`

`LHS=0`

`LHS=RHS`

Answer 2

`color(green)(=> 0 = R H S`

Explanation:

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

`cos ^2 x + sin ^2 x + tan ^2 x - (1/cos^2 x)`, rearranging terms

`cos^2 x + sin ^2 = 1`, identity

`sec x -= 1/cos x`, identity

`:. cancel(cos^2 x + sin^2 x)^color(red)(1) + tan^2 x - sec^2 x`

`1 + tan^2 x = sec ^2x`, identity

`=> cancel(1 + tan^2 x )- cancel(sec ^2 x )= 0`

`color(green)(=> R H S`