Evaluate 1+tan(theta)^2=sec(theta)^2

$1 + \tan^{2} (θ) = \sec^{2} (θ)$

Simplify $1 + \tan^{2} (θ)$ .

Rearrange terms.

$\tan^{2} (θ) + 1 = \sec^{2} (θ)$

Apply pythagorean identity.

$\sec^{2} (θ) = \sec^{2} (θ)$

For the two functions to be equal, the arguments of each must be equal.

$θ = θ$

Move all terms containing $θ$ to the left side of the equation.

Subtract $θ$ from both sides of the equation.

$θ - θ = 0$

Subtract $θ$ from $θ$ .

$0 = 0$

Since $0 = 0$ , the equation will always be true.

Always true

The result can be shown in multiple forms.

Always true

Interval Notation:

$(- \infty, \infty)$