Question

Bah, I'll ask this one with that equation thing.

Answer 1

s\[Simplify: \sec^2\theta - 1/\sec^2 \theta \]

Answer 2

sec cancle out right?

Answer 3

just for kicks is it

1. (sec^2 - 1)

2. sec^2 - (1/sec^2) ?

Answer 4

Well our teachers gave us the answers, I just don't know how to solve it.. but the answer is \[\sin ^2\theta \]

Answer 5

nvm my question

this is gonna take a second not used to the equation thing but it is just a trig identity

Answer 6

Okay, take your time o;

Answer 7

\[(\sec ^2 - 1) / \sec ^2\]
\[= \tan ^2 / \sec^2\]
\[= (\sin^2/\cos^2)/\sec^2\]
\[= (\sin^2/\cos^2)/(1/\cos^2)\]
\[= (\sin^2 * \cos^2 / \cos^2 )\]

Answer 8

which equals
\[\sin^2\]

i used \[1 + \tan^2 = \sec^2\]
and changed it to
\[\tan^2 = \sec^2 - 1\]
and exchanged sec - 1 for tan in the problem

Answer 9

In that second to last step, you can just drop the 1? o:

Answer 10

no, what i did was invert and multiply

because it was sin/cos, divided 1/cos.

invert and multiply you get

sin/cos * cos/1, or just sin/cos * cos, which is sin*cos/cos, and cos/cos is just 1

so you are left with sin

Answer 11

that's all sin^2 and cos^2 of course lol

Answer 12

Oh alright, thank you C:

Answer 13

Is the question \[(\sec ^{2}\theta -1)/\sec ^{2}theta\]

or \[\sec ^{2}\theta - 1/\sec ^{2}\theta\]

Answer 14

Just how I typed it, which is the second one. It's been solved

Answer 15

sin(x)^2

Answer 16

Hm?

Answer 17

If it's the second one then I'm afraid that solution was incorrect because the you have to find the common denominator first

Answer 18

Well the right answer was the result of it..?

Answer 19

^^ im trying to find a work around for the second one lol

Answer 20

and just for kicks, the way it is written whereever you are finding it, is it like this