Question

I'm trying to verify this expression:
1 + sec^2 x sin^2 x = sec^2 x

This is what I've done:

sec^2 x
=1+tan^2 x
=sin^2 x +cos^2 x +tan^2 x
=(sin^2 x cosx/cosx)+(cos^2x sinx/sinx)+(sinx/cosx)
=sinx[sinx cosx/cosx)+(cos^2 x/sinx)+(1/cosx)

I'm pretty sure I'm doing this all wrong! Please help!

Answer 1

i made a mistake before so now is the correct one
secx=1/cosx and sec^2xsin^2x=tan^2x so 1+tan^2x=sec^2x

Answer 2

Sorry I left for a few minutes I have to match the right side to the left side

Answer 3

how about changing \(\sec^2 x\) to \(\displaystyle \frac{1}{\cos^2 x}\)?

Answer 4

@mikoza already gave the solution.

Answer 5

Isn't that matching the left side to the right?

Answer 6

yes it is.

Answer 7

I'm supposed to change the right side to match the left. I know the other one is easier, but I'm having trouble matching the right to left

Answer 8

\[\sec^2x=1+\tan^2x=1+\frac{\sin^2x}{\cos^2x}=1+\frac{1}{\cos^2x} \sin^2x=1+\sec^2x \sin^2 x\]

Answer 9

left-to-right is not different from right-to-left.

Answer 10

sec^2x=1/cos^2x=(cos^2x+sin^2x)/cos^2x=cos^2x/cos^2x+sin^2x/cos^2x=1+sin^2xsec^2x

Answer 11

Oh! Thank you both so much! :)