If \( \sec^2 \theta + \tan^2 \theta = 3 \), then the angle \( \theta \) is equal to?

We're given the equation sec²θ + tan²θ = 3, and we need to solve for θ. One way to approach this is to use the trigonometric identity: tan²θ + 1 = sec²θ Substituting this into the given equation, we get: tan²θ + 1 + tan²θ = 3 Simplifying this equation, we get: 2tan²θ = 2 Dividing both sides by 2, we get: tan²θ = 1 Taking the square root of both sides, we get: tanθ = ±1 This means that θ must be one of the angles whose tangent is ±1. These angles are 45º and 225º (or -135º), since tangent is positive in the first and third quadrants, and negative in the second and fourth quadrants. However, we need to check whether these values of θ satisfy the original equation. Let's start with θ = 45º: sec²45º + tan²45º = 2 + 1 = 3 So this value of θ does satisfy the equation, and therefore it is the solution. Therefore, the angle θ is 45º. So the answer is (c) 45º.