Question

- tan2x + sec2x = 1

Answer 1

Is this supposed to be \[-tan^2x + sec^2x = 1\] or are they correct as written (if squared it is a considerably easier problem, and I'm not even sure the correct the other way)

Answer 2

what u wrote is what it is supposed 2 b

Answer 3

Okay, here you just need the Pythagorean trig identities. Called as such because they are of the form a^2 + b^2 = c^2.
There is one for sin and cos \[cos^2x + sin^2x = 1\]
One for sec and tan \[1+ tan^2x = sec^2x\]
And one for csc and cot \[1+cot^2x = csc^2x\]
We are interested in the second one for this equation.
Do you see how to do it from there or still need help?

Answer 4

i still need some help?

Answer 5

sorry bout the question mark

Answer 6

Okay, try re-ordering the left side so that the sec^2(x) comes first. So:
\[sec^2x - tan^2x =1\]

Answer 7

is that the answer?

Answer 8

If you're trying to show that the original equation is in fact 1, then not quite. If you're just trying to simplify the original equation then yes.

Answer 9

the instructions are to verify the trigonometric equation by substituting identities

Answer 10

Okay, then they probably just want you to show that the original equation can be reordered to be \[sec^2x = 1+ tan^2x\]
since that is the "standard form" for any of the Pythagorean identities.

Answer 11

So, we have it simplified as \[sec^2x - tan^2x = 1\] what do we need to do to get tan^2(x) to the other side?

Answer 12

add it to both sides

Answer 13

yep. And does that match the standard form for the sec/tan Pythagorean identity?

Answer 14

yes. so that is my final answer rite?

Answer 15

yep.

Answer 16

thank u soooooo much!!!

Answer 17

You're welcome. Do you have a reference sheet with different trig identities on it?

Answer 18

yes i just looked it up and found one.