Question

Please help me identify the identity used in this precalc/trig problem! Click attached

Answer 1

Did you guess or did you do actual steps?

Answer 2

We need an identity that relates sec^2(x) and tan^2(x)

We know that\[\sin ^{2}(x) + \cos ^{2}(x) =1\]

Divide both sides by cos^2(x) and we are left with
\[\tan^{2}(x) + 1 = \sec^{2}(x)\]

And this identity relates the two trig functions we wanted

Answer 3

Thank you guys! So from your instruction @jheidelk, my selection is correct.

Answer 4

Funny thing is, I don't use any of those substitutions in my solutions

Answer 5

I can show you my steps

Answer 6

Hmm, really? I'd like to see how you worked it out please

Answer 7

Actually @IloveCharlie sort of saw how I approach mine

Answer 8

You already saw some of the way I approach these, but I'll post it on here (or at least try to)

Answer 9

Thank you Hero!

Answer 10

\(\tan^2 x + \sec^2 x = 3\)

\(\large\frac{\sin^2 x}{\cos^2 x} + \frac{1}{\cos^2 x} = 3\frac{\cos^2x}{cos^2x}\)

This is how I start off. Then I multiply both sides by cos^2x to get:
\(\sin^2x + 1 = 3\cos^2x\)

Answer 11

Okay, I admit, I do use one substitution, but nothing more than \(\sin^2x + \cos^2x = 1\)

Answer 12

Which in the next step I get:

\(\sin^2x + \sin^2 x + \cos^2 x = 3 \cos^2x\)

Now I combine sin^2x and then subtract cos^2x from both sides to get:
\(2\sin^2x = 2\cos^2x\)

Then divide both sides by two to get:

\(\sin^2x = \cos^2x\)

Answer 13

I make one last substitution, but that's just a variation of \(\sin^2x + \cos^2x = 1\):

\(\sin^2x - \cos^2x = 0\) and
\(\sin^2x - (1-\sin^2x) = 0\)

And you can probably see where I'm going with this by now.

Answer 14

\(\sin^2x + \sin^2x - 1 = 0\)

\(2\sin^2x = 1\)
\(\sin^2x = \frac{1}{2}\)

\(\sin x = \pm \sqrt{\frac{1}{2}}\)

Answer 15

\(x = \sin^{-1} (\pm \sqrt{\frac{1}{2}})\)

Answer 16

You're awesome. Thank you for your step by step method :)

Answer 17

Which is basically just

\(x =\large \pm \frac{\pi}{4}\)

Answer 18

@IloveCharlie enjoys my step-by step methods :P

Answer 19

I feel as though @jheidelk deserves a middle as well. Would you mind giving him one? I would but I can't :(

Answer 20

*medal