Question

I need help proving this trigonometric identity step-by-step? (Will give medal and fan)

tan^2 x - sin^2 x = tan^2 x sin^2 x

I'm given two columns: Calculation and Reason. I was given the reasons (which I will list in order in the comments), but I need to list my calculations according to the reasons.

Answer 1

Reasons:

1. This follows the definition of tan x = sin x / cos x
2. This follows from factoring out sin^2 x
3. This follows by the definition of sec x = 1/cos x
4. This follows from the Pythagorean identity sec^2 x - 1 = tan^2 x

Answer 2

@whpalmer4 ?

Answer 3

oops last step must be (sin^2x * sin^2x) / cos^2x

Answer 4

then we will get tan^2x *sin^2x

Answer 5

Have you understood ?

Answer 6

Can you explain to me how those steps refer to the reasons? Like which step corresponds with which reason? Sorry or the late response

Answer 7

*for

Answer 8

At left hand side we replace tan^2x by sec^2x - 1 okay ?

Answer 9

But the first reason says I'm supposed to use the identity tanx = sinx/cosx, does that go along with that reason?

Answer 10

this is trigonometric formula ..

Answer 11

okay wait

Answer 12

alright thank you

Answer 13

Okay we are solve using this tanx = sinx/cosx So tan^2x = sin^2x / cos^x
So replace tan^2x by sin^2x / cos^2x at left hand side ,, what do you get ?

Answer 14

sin^2 x / cos^2 x - sin^2 x ?

Answer 15

correct !! now take LCM

Answer 16

what's LCM?

Answer 17

The truth is, different people have different methods for figuring out trigonmetric identities. In this case, while some would choose to take the route of converting to \(\sec^2x\) to finish solving, I am perfectly fine with producing the following set of steps:

\(\dfrac{\sin^2x}{\cos^2x} - \sin^2x\)

Factor out \(\sin^2x\) to get:

\(\sin^2x\left(\dfrac{1}{\cos^2x} - 1\right)\)


Re-write 1 as \(\dfrac{\cos^2x}{\cos^2x}\):

\(\sin^2x\left(\dfrac{1}{\cos^2x} - \dfrac{\cos^2x}{\cos^2x}\right)\)

Combine fractions:
\(\sin^2x\left(\dfrac{1-\cos^2x}{\cos^2x}\right)\)

Replace \(1 - \cos^2x\) with \(\sin^2x\):
\(\sin^2x\left(\dfrac{\sin^2x}{\cos^2x}\right)\)

\(\sin^2x \tan^2x\)

Answer 18

Is that the finished proof? Thank you so much!

Answer 19

No, that still isn't what you're looking for, but I can demonstrate how they wanted you to do it. I was just expressing a point that they are forcing you to do it "only one way" which I don't exactly agree with.

Answer 20

What they wanted you to do is this:

\(\dfrac{\sin^2x}{\cos^2x} - \sin^2x\)

Factor out \(\sin^2x\):
\(\sin^2x\left(\dfrac{1}{\cos^2x} - 1\right)\)

Observe that \(\dfrac{1}{\cos^2x} = \dfrac{1}{\cos(x)}\dot\ \dfrac{1}{\cos(x)}\)
\(\sin^2x\left(\dfrac{1}{\cos(x)}\dot\ \dfrac{1}{\cos(x)} - 1\right)\)

Recall that \(\dfrac{1}{\cos(x)} = \sec(x)\):
\(\sin^2x\left(\sec(x)\sec(x)- 1\right)\)

\(\sin^2x\left(\sec^2x - 1\right)\)

And finally by Pythagorean Identity:
\(\sin^2x\tan^2x\)

Answer 21

Something like that