Prove this identity: $\cos x(\sec x-\cos x)=\sin^2x$

Question

I have been trying to prove this identity for about 25 minutes and I can't figure it out. Can you help me please?

$$\cos x(\sec x-\cos x)=\sin^2x$$

Answer 1

See that $\sec x-\cos x=\frac{(1-\cos^2x)}{\cos x}=\frac{\sin^2x}{\cos x}$

Now if you multiply it by $\cos x$ then you get your answer $\sin^2x$.

Answer 2

Note that by the definition of the secant: $$\color{blue}{\sec x = \frac{1}{\cos(x)}}$$ So rewriting $\sec x$ as $\frac{1}{\cos(x)}$ in your question, we have:

$$\cos x\left(\color{blue}{\frac{1}{\cos x}}-\cos x\right)=\sin^2x$$

By the distributive property we can multiply the $\cos x$ in the sum (or difference), then we'll get:

$$1-\cos^2 x = \sin^2x$$

This is true because of the identity:

$$\sin^2 x + \cos^2x = 1$$

Answer 3

Cos(x)(sec(x)-cos(x))=cos(x)(1/cos(x)-cos(x))=cos(x)((1-cos^2(x))/cos(x))=cos(x)(sin^2(x)/cos(x))=sin^2(x). Where sec(x)=1/cos(x) and cos^2(x)+sin^2(x)=1.