Question

is Cos^2x(1+tan^2x)=1
an identity

Answer 1

\[\cos^2x\left(1+\tan^2x\right)=\cos^2x\left(1+\frac{\sin^2x}{\cos^2x}\right)\]What happens when you distribute the \(\cos^2x\) term?

Answer 2

What did you just do??

Answer 3

Why is it an identity?

Answer 4

I'm just using the definition of \(\tan x\), which is the ratio of \(\sin x\) to \(\cos x\).

Answer 5

does it have to be simplified into one of the trig identities?

Answer 6

That's what you need to do after distributing, yes.
\[\cos^2x\left(1+\frac{\sin^2x}{\cos^2x}\right)=\cos^2x+\sin^2x=\cdots\]which should look familiar to you.

Answer 7

I'm not sure how you got that

Answer 8

This is the distributive property of multiplication:\[a(b+c)=ab+ac\]with \(a=\cos^2x\), \(b=1\) and \(c=\tan^2x=\dfrac{\sin^2x}{\cos^2x}\).

More explicitly,
\[\cos^2x\left(1+\frac{\sin^2x}{\cos^2x}\right)=\cos^2x+\frac{\sin^2x\cos^2x}{\cos^2x}=\cos^2x+\sin^2x=\cdots\]

Answer 9

cos^2x+sin^2x=1

Answer 10

@HolsterEmission

Answer 11

Right, so the original equation is indeed an identity.

Answer 12

This is the pythagorean identity

Answer 13

Yup. A lot of these "prove this is an identity" questions reduce to this fundamental identity.

Answer 14

thanks so much

Answer 15

yw