Question

Match equation on right; Trogonometry

Answer 1

\[1+\sec^2x \sin^2x=\sec^2x\]

Answer 2

Nice seeing you again, but I'm not permitted to change the right side of the equation

Answer 3

sorry,identity was wrong, let me retry.

Answer 4

What's up here? Prove that this is true?

Answer 5

Yup

Answer 6

Change the sec^2 on the left to a function of cos.

Answer 7

@NoelGreco would that become 1/cos^2?

Answer 8

yes
. now sin^2/cos^2, and you've got a Pythagorean identity.

Answer 9

okay!
we know that
\[\sin^{2}x = 1 - \cos^{2}x\], so lets alter it up a bit!
\[1 + \sec^{2}x(1-\cos^{2}x)\]
\[1 + \sec^{2}x - \sec^{2}xcos^{2}x\]
and because \[secx= \frac{1}{cosx}\] we know that multiplying secant and cosine will always give you 1
\[1 - 1+ \sec^{2}x = \sec^{2}x\]

Answer 10

Good job.

Answer 11

why don't we turn sec^2 sin^2 = sin^2/cos^2 = tan^2
so the whole thing is 1+ tan^2 = sec^2?

Answer 12

That's it.

Answer 13

yes, I think so, smarties!! am i wrong?

Answer 14

Nope! That's the fun with trig identities, you can express them millions of different ways, but still be correct.

Answer 15

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

Answer 16

Yep.. I like he way @Loser66 worked it out. My result was the same.

Answer 17

I actually didn't know secane times cosine equaled 1 =p Thank you, ive gotta write that down. Thank you all for comfirming this

Answer 18

yes because \[\sec(x)=\frac{1}{\cos(x)}\] and \[\sec(x)*\cos(x)= \frac{1}{cosx}* \frac{\cos(x)}{1}=1\]