Question

Choose the alternative that best completes the statement or answers the question:
cosx/tan^2x-cosx/sin^2x

A) -cosx B) -sinx C) cosx D)sinx

Answer 1

$$
\cfrac{\cos x}{\tan^2x}-\cfrac{\cos x}{\sin^2x}\\
=\cfrac{\cos x\times\cos^2 x}{\sin^2x}-\cfrac{\cos x}{\sin^2x}\\
=\cfrac{\cos x\times\cos^2 x-\cos x}{\sin^2 x}\\
=\cfrac{\cos x(\cos^2x-1)}{\sin^2x}\\
=\cfrac{\cos x \sin^2x}{\sin^2x}\\
=\cos x
$$
Make sense?

Answer 2

isn't it a

Answer 3

Yes, you are right. Thanks for catching that!
$$
\cfrac{\cos x}{\tan^2x}-\cfrac{\cos x}{\sin^2x}\\
=\cfrac{\cos x\times\cos^2 x}{\sin^2x}-\cfrac{\cos x}{\sin^2x}\\
=\cfrac{\cos x\times\cos^2 x-\cos x}{\sin^2 x}\\
=\cfrac{\cos x(\cos^2x-1)}{\sin^2x}\\
=-\cfrac{\cos x \sin^2x}{\sin^2x}\\
=-\cos x
$$

Because
$$
\sin^2x+cos^2x=1\\
\sin^2x=-cos^2x+1\\
\sin^2x=1-cos^2x\\
\implies \cos^2x-1=-\sin^2 x\\
$$