Question

in exercises 9 and 10, verify the identity. 9. $dfrac{1 - cos^{2}\theta}{sec^{2}\theta} - sin^{2}\theta = -sin^{4}\theta$

Explanation:

Step1: Use trigonometric identities

We know that \(1 - \cos^{2}\theta=\sin^{2}\theta\) and \(\sec^{2}\theta=\frac{1}{\cos^{2}\theta}\).
So, \(\frac{1 - \cos^{2}\theta}{\sec^{2}\theta}=\sin^{2}\theta\times\cos^{2}\theta\).

Step2: Substitute into the left - hand side

The left - hand side of the equation \(\frac{1 - \cos^{2}\theta}{\sec^{2}\theta}-\sin^{2}\theta\) becomes \(\sin^{2}\theta\cos^{2}\theta-\sin^{2}\theta\).
Factor out \(\sin^{2}\theta\): \(\sin^{2}\theta(\cos^{2}\theta - 1)\).
Since \(\cos^{2}\theta-1=-\sin^{2}\theta\), then \(\sin^{2}\theta(\cos^{2}\theta - 1)=-\sin^{4}\theta\).

Answer:

The identity \(\frac{1 - \cos^{2}\theta}{\sec^{2}\theta}-\sin^{2}\theta=-\sin^{4}\theta\) is verified.