Question

proving identities:
see picture below

Answer 1

@Callisto

Answer 2

\[\frac{1-\tan}{\sec}+\frac{\sec}{\tan}\]\[=\frac{\tan(1-\tan)+\sec^2}{\sec\tan}\]
Note: \(\tan^2-\sec^2 = 1\)

Answer 3

I got stuck with

tan - 1 / sec tan

what do i do next? @Callisto

Answer 4

Where did you get tan - 1 / sec tan ?

Answer 5

there ^

Answer 6

did i do it correctly?

Answer 7

Hmm, hold on.

Answer 8

ok

Answer 9

Sorry for the mistake made earlier.
It should be 1 = sec^2 - tan^2
The way to get this identity is as follows:
\[\sin^2+\cos^2 = 1\]\[\frac{\sin^2}{\cos^2}+\frac{\cos^2}{\cos^2}=\frac{1}{\cos^2}\]\[\tan^2 + 1 = \sec^2\]
I made a mistake when I rearranged the terms in this identity. Sorry again!

Answer 10

it's fine.

thanks i got it now :)

Answer 11

You're welcome :)