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