Answer: To show that the function has exactly one zero in the given interval, we need to show that the function changes sign exactly once on the interval.
First, we can see that the function is continuous on the interval (0, pi/2) as it is a polynomial of trigonometric functions.
Next, we can evaluate the function at the endpoints of the interval:
R(0) = sec^2(0) - cos(0) - 1 = 1 - 1 - 1 = -1
R(pi/2) = sec^2(pi/2) - cos(pi) - 1 = undefined
We can see that R(0) is negative, and since the function is continuous on the interval, by the Intermediate Value Theorem, the function must pass through zero at some point in the interval.
To show that it passes through zero only once, we can take the derivative of the function:
R'(theta) = 2sec^2(theta)sin(theta) + 2sin(2theta)
We can see that R'(theta) is positive for all theta in the interval (0, pi/2), as sec^2(theta) and sin(theta) are positive and sin(2theta) is non-negative. Therefore, the function R(theta) is strictly increasing on the interval, and can cross the x-axis at most once.
Thus, we have shown that the function R(theta) has exactly one zero in the interval (0, pi/2).
Explanation: