couple trig identities

[corsec]

hi I need some help with a couple of trig identies that I need to prove

t = theta

the first one is tan^2t which should be (sin^2t)/(cos^2t) right? which also equals

(1-cos[2t])/(1+cos[2t])

but the answers i'm given are all 1 +or - sec^2t or csc^2 or the wonderful none of the above! hate them

same thing with the second one it's find cos^2t with answers 1 + or - sec^2 or 1 + or - sin^2 etc. or of course none of the above

I know on the second one it equals

(1+cos[2t])/2 but as far as sec or sin I'm not sure

 

[soroban]

Hello, corsec!

It doesn't sound like you have to prove anything
. . . just decide which one it equals.

And these are so simple, you'll kick yourself!


Of course, I guessing . . . Does the first problem look like this?

\(\displaystyle 1)\;\;\tan^2\theta\:=\)

\(\displaystyle a)\;1\,-\,\sec^2\theta\;\;\;b)\;1\,+\,\sec^2\theta\;\;\;c)\;1\,+\,\csc^2\theta\;\;\;d)\;1\,-\,\csc^2\theta\;\;\;e)\;\text{none of the above}\)

You're expected to know the identity: \(\displaystyle \sec^2\theta\:=\:\tan^2\theta\,+\,1\)
\(\displaystyle \;\;\) which can be written: \(\displaystyle \,\tan^2\theta\:=\;\sec^2\theta\,-\,1\)

So, the answer is \(\displaystyle \,(e)\;\text{none of the above}\)


And is the second one something like this?

\(\displaystyle 2)\;\;\cos^2\theta\;=\)

\(\displaystyle a)\;1\,+\,\sec^2\theta\;\;\;b)\;1\,-\,\sec^2\theta\;\;\;c)\;1\,+\,\sin^2\theta\;\;\;d)\;1\,-\,\sin^2\theta\;\;\;e)\;\text{none of the above}\)

I'm sure you know the identity: \(\displaystyle \sin^2\theta\,+\,\cos^2\theta\:=\:1\) . . . right?

Then: \(\displaystyle \,\cos^2\theta\;=\;1\,-\,\sin^2\theta\;\) . . . answer (d)

 

[corsec]

your right those are so simple, I guess it was one of those cases of trying to overanalyze the problem! Thank you for the insight. Simple algebra I can't believe it