Question

@safl;jk tan^2 x=sec^2x-1

Answer 1

true :)

Answer 2

use sin^2x+cos^2x=1 , divide all by cos^2x

Answer 3

hahaha yaaaa
We need to show @safl;jk how this is an identity
What i would do is simplify the right hand side and show that it equals the left hand side

Answer 4

The following are 2 trig identities that will help you solve the question:
\[ \sec^2x=\frac{1}{\cos^2x}\]
\[ \sin^2x + \cos^2x =1\]

Answer 5

So lets only focus on the right hand side of the equation which is \( \sec^2x-1\)
Right Hand Side:
\[ \sec^2x -1\]
\[ \frac{1}{\cos^2x}-1\]
\[\frac{1}{\cos^2x} - \frac{ \cos ^2x }{\cos^2x}\]
\[ \frac{1-\cos^2x}{\cos^2x}\]
Now we know the trig identity that sin^2x+cos^2x =1
Using some algebra we can rearrange the equation to :
1-cos^2x =sin^2x
Therefore we can replace 1-cos^2x with sin^2x in our equation above
\[ \frac{1-\cos^2x}{\cos^2x}=\frac{\sin^2x}{\cos^2x} =\tan^x\]
SO the right hand side of the equation is tan^2x and so the right hand side of the equation equals the left hand side of the equation and therefore we can agree that this equation is true

Answer 6

@safl;jk Come check this out