Question

Will reward medal (if told how). Prove the identity: tan^2(x) - sin^2(x) = tan^2(x) * sin^2(x) using the following "reasons". Step 1: "This follows the definition of tanx=sinx/cosx." Step 2: "This follows from factoring out sin^2(x)." Step 3: "This follows by the definition sec x = 1/cos x." Step 4: "This follows from the Pythagorean identity sec^2(x) - 1 = tan^2(x)"

Answer 1

I'm pretty sure I have step 1 down, but I can't seem to get step 2. Any help would be greatly appreciated. :)

Answer 2

For step 1, you re-wrote the LHS as:

\[\frac{ \sin^2(x) }{ \cos^2(x) } - \sin^2(x)\]
right?

Now factor out the \(\sin^2(x)\) in that expression. What do you get now on the LHS?

Answer 3

Sorry, but I can't see the math that you're writing. It's saying "[Math processing error]". I rewrote the LHS as (sin^2(x)/cos^2x) - sin^2(x)

Answer 4

weird.... ok, that's what I wrote. :) No, if you factor out the sin^2(x) from that, what do you get?

Answer 5

E.g., your "reasons" are basically telling you exactly what to do in each step. When you are finished, you will have gone from the LHS to the RHS.

Answer 6

That's what I'm a bit confused on. Would the sin values cancel out, or do you have to find the common denominator?

Answer 7

No, neither. Just factor. If I have:

a/b-a

that is really a/b - a/1 right?

so I can factor out like so:

a(1/b - 1)

Answer 8

So sin^2x(1/cos^2x - 1)?

Answer 9

right - good! Do you see where to go next?

Answer 10

Yes, I believe so. Thank you so much! How do I give you a medal? (sorry, I'm new to the website)

Answer 11

I see you figured it out. :) Thank you, and you're welcome. happy to help :)