Question

i need help to verify (tan^2x)/(1-cosx) = secx+sec^2x

Answer 1

the trig identity for tangent ^2 is sec^-1

Answer 2

sec^2-1

Answer 3

okay yea that makes sense so far.

Answer 4

I'm not sure if the trig identity for tangent ^2 is sec^2x-1..would it be \(tan^{2}x=\Large \frac{sin^2x}{cos^2x}\)?

Answer 5

well the pythagorean identity for tangent^2 is sec^2x-1 at least that is what my teacher gave me idk if i can do it the other way. but that might make it easier.

Answer 6

brb, i'll try to ask someone

Answer 7

okie dokes

Answer 8

Ok, it's correct

Answer 9

I apologize for taking too long

Answer 10

okay cool its fine what would i do from there.

Answer 11

\[\Large \frac{\tan^2(x)}{1-\cos(x)} = \sec(x) + \sec^2(x)\]
\[\Large \frac{\sec^2(x)-1}{1-\cos(x)} = \sec(x) + \sec^2(x)\]
\[\Large \frac{\frac{1}{\cos^2(x)}-1}{1-\cos(x)} = \sec(x) + \sec^2(x)\]
\[\Large \frac{\cos^2(x)}{\cos^2(x)}*\frac{\frac{1}{\cos^2(x)}-1}{1-\cos(x)} = \sec(x) + \sec^2(x)\]
\[\Large \frac{\cos^2(x)(\frac{1}{\cos^2(x)}-1)}{\cos^2(x)(1-\cos(x))} = \sec(x) + \sec^2(x)\]
\[\Large \frac{1-\cos^2(x)}{\cos^2(x)(1-\cos(x))} = \sec(x) + \sec^2(x)\]
\[\Large \frac{(1-\cos(x))(1+\cos(x))}{\cos^2(x)(1-\cos(x))} = \sec(x) + \sec^2(x)\]
Hopefully you see what to do from here

Answer 12

i kind of do but im not to good at this. could you help me finish i know you would use reciprocal identities but..