Question

tanxcotz-cos^2x=sin^2x

Answer 1

Ah hA. Thought so. Now, can you see how what I did the first time would help here?

Answer 2

\[\tan x \cot x-\cos^2x=\sin^2x \implies \\
\frac{\sin x}{\cos x} \frac{\cos x}{\sin x} -\cos^2x=\sin^2x\implies ??\]So just by turning it into sin and cos I have something easier.

Answer 3

i did but i cloudnt get the same answer as the other side

Answer 4

Well, if you simplify that multiplied part, what does it become?

Answer 5

huh

Answer 6

Cancel the cos/cos and sin/sin parts and what do you get?

Answer 7

\[\frac{\sin x}{\cos x} \frac{\cos x}{\sin x} -\cos^2x=\sin^2x\implies \\
\frac{\sin x \cos x}{\sin x \cos x} -\cos^2x=\sin^2x\]

Answer 8

Look at just this part for now:
\[\frac{\sin x \cos x}{\sin x \cos x}=???\]

Answer 9

do i times the bottom to the top one??

Answer 10

Remember your rules for simpifying terms in fractions.
\[\frac{5}{5}=\frac{a}{a}=\frac{m\times n}{m\times n}=\frac{\sin x \cos x}{\sin x \cos x}=?\]

Answer 11

no thats not what i'm doing i have to make both sides equal to each other

Answer 12

Actually, this is what you are doing. The rules of fractions apply in trig.
We had two fractions and we multiplied them. They were originally tan and cot. But we rewote them. Let me draw it and you might remember doing this.