Question

Please Help!

Answer 1

ok how

Answer 2

Choice B is NOT an identity. If you plugged in theta = pi/4, you'll find that the left side turns into 4. That's one way you can check to see if it's an identity or not.

Answer 3

I apologize ! I was incorrect

Answer 4

which one?

Answer 5

If you plug theta = pi/4 into choice B, this is what happens

\[\Large \sec^2\left(\theta\right)+\csc^2\left(\theta\right) = 1\]

\[\Large \frac{1}{\cos^2\left(\theta\right)}+\frac{1}{\sin^2\left(\theta\right)} = 1\]

\[\Large \frac{1}{\cos^2\left(\pi/4\right)}+\frac{1}{\sin^2\left(\pi/4\right)} = 1\]

\[\Large \frac{1}{\left(\cos\left(\pi/4\right)\right)^2}+\frac{1}{\left(\sin\left(\pi/4\right)\right)^2} = 1\]

\[\Large \frac{1}{\left(1/\sqrt{2}\right)^2}+\frac{1}{\left(1/\sqrt{2}\right)^2} = 1\]

\[\Large \frac{1}{1/2}+\frac{1}{1/2} = 1\]

\[\Large 2+2 = 1\]

\[\Large 4 = 1\]

Since the equation is not true, this means we do NOT have an identity.

Answer 6

o

Answer 7

which one would work?

Answer 8

Choice A doesn't work either


\[\Large \cot\left(\theta\right)+\tan\left(\theta\right) = 1\]

\[\Large \frac{1}{\tan\left(\theta\right)}+\tan\left(\theta\right) = 1\]

\[\Large \frac{1}{\tan\left(\frac{\pi}{4}\right)}+\tan\left(\frac{\pi}{4}\right) = 1 \ ... \ \text{Plug in } \theta = \frac{\pi}{4}\]

\[\Large \frac{1}{1}+1 = 1\]

\[\Large 1+1 = 1\]

\[\Large 2 = 1\]


So choice A isn't an identity either

Answer 9

so choice C :)

Answer 10

Checking choice D

\[\Large \cos^2\left(\theta\right) = \sec^2\left(\theta\right)\times\tan^2\left(\theta\right)\]

\[\Large \cos^2\left(\frac{\pi}{4}\right) = \sec^2\left(\frac{\pi}{4}\right)\times\tan^2\left(\frac{\pi}{4}\right)\]

\[\Large \left(\cos\left(\frac{\pi}{4}\right)\right)^2 = \left(\sec\left(\frac{\pi}{4}\right)\right)^2\times\left(\tan\left(\frac{\pi}{4}\right)\right)^2\]

\[\Large \left(\frac{1}{\sqrt{2}}\right)^2 = \left(\sqrt{2}\right)^2\times\left(1\right)^2\]


\[\Large \frac{1}{2} = 2\]


which is false, so choice D isn't an identity either

Answer 11

`so choice C :)`
yes that's correct @Doge_

Answer 12

btw you don't have to pick pi/4. I just picked something between 0 and pi/2 in quadrant 1. You can pick any value for theta. I picked on pi/4 because it's found on the unit circle.

Answer 13

Choice C

\[\Large \tan^2\left(\theta\right) = \sin^2\left(\theta\right) \times \sec^2\left(\theta\right)\]

\[\Large \tan^2\left(\theta\right) = \sin^2\left(\theta\right) \times \frac{1}{\cos^2\left(\theta\right)}\]

\[\Large \tan^2\left(\theta\right) = \frac{\sin^2\left(\theta\right)}{1} \times \frac{1}{\cos^2\left(\theta\right)}\]

\[\Large \tan^2\left(\theta\right) = \frac{\sin^2\left(\theta\right)\times 1}{1 \times\cos^2\left(\theta\right)}\]

\[\Large \tan^2\left(\theta\right) = \frac{\sin^2\left(\theta\right)}{\cos^2\left(\theta\right)}\]

\[\Large \tan^2\left(\theta\right) = \tan^2\left(\theta\right)\]

So the identity is confirmed

Answer 14

ok
doge_ got suspended but im him basement roommate and he said question closed