Question

staking with vocabulary (continued) date

1. \\(\frac{2\sec^2 x - 2\tan^2 x}{\tan(-x)\cos(-x)}\\)
2. \\(\frac{-sin\left(\frac{\pi}{2} - \theta\

ight)}{\sec(-\theta)} - \sin^2 \theta\\)

in exercises 9 and 10, verify the identity.

1. \\(\frac{1 - \cos^2 \theta}{\sec^2 \theta} - \sin^2 \theta = -\sin^4 \theta\\)
2. \\(\csc x + \cot x = \frac{\sin x}{1 - \cos x}\\)

Explanation:

Step1: Use the Pythagorean identity \(1 - \cos^{2}\theta=\sin^{2}\theta\) and \(\sec^{2}\theta=\frac{1}{\cos^{2}\theta}\)

Substitute \(1 - \cos^{2}\theta=\sin^{2}\theta\) into \(\frac{1 - \cos^{2}\theta}{\sec^{2}\theta}-\sin^{2}\theta\). We get \(\frac{\sin^{2}\theta}{\sec^{2}\theta}-\sin^{2}\theta\). Since \(\sec^{2}\theta=\frac{1}{\cos^{2}\theta}\), then \(\frac{\sin^{2}\theta}{\sec^{2}\theta}=\sin^{2}\theta\cos^{2}\theta\). So the expression becomes \(\sin^{2}\theta\cos^{2}\theta-\sin^{2}\theta\)

Step2: Factor out \(\sin^{2}\theta\)

Factor out \(\sin^{2}\theta\) from \(\sin^{2}\theta\cos^{2}\theta-\sin^{2}\theta\). We have \(\sin^{2}\theta(\cos^{2}\theta - 1)\)

Step3: Use the Pythagorean identity \(\cos^{2}\theta=1 - \sin^{2}\theta\)

Substitute \(\cos^{2}\theta = 1-\sin^{2}\theta\) into \(\sin^{2}\theta(\cos^{2}\theta - 1)\). Then \(\sin^{2}\theta((1 - \sin^{2}\theta)-1)=\sin^{2}\theta(-\sin^{2}\theta)=-\sin^{4}\theta\)

Answer:

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