Answer
$$\sin^2(-\theta)+\tan^2(-\theta)+\cos^2(-\theta)=\sec^2\theta$$
Work Step by Step
$$A=\sin^2(-\theta)+\tan^2(-\theta)+\cos^2(-\theta)$$
$$A=[\sin(-\theta)]^2+[\tan(-\theta)]^2+[\cos(-\theta)]^2$$
- Negative-Angle Identities:
$$\sin(-\theta)=-\sin\theta\hspace{1cm}\tan(-\theta)=-\tan\theta\hspace{1cm}\cos(-\theta)=\cos\theta$$
However,
$$[\sin(-\theta)]^2=[-\sin\theta]^2=\sin^2\theta$$
$$[\tan(-\theta)]^2=[-\tan\theta]^2=\tan^2\theta$$
$$[\cos(-\theta)]^2=[\cos\theta]^2=\cos^2\theta$$
(for $(-A)^2=A^2$)
Therefore,
$$A=\sin^2\theta+\tan^2\theta+\cos^2\theta$$
$$A=(\sin^2\theta+\cos^2\theta)+\tan^2\theta$$
- Pythagorean Identity:
$$\sin^2\theta+\cos^2\theta=1$$
So, $$A=1+\tan^2\theta$$
- Pythagorean Identity:
$$\tan^2\theta+1=\sec^2\theta$$
which means,
$$A=\sec^2\theta$$
