Question

Which of the following is a trigonometric identity?

Answer 1

\(\sin^2\theta\tan^2\theta=\tan^2\theta+\sin^2\theta\)
\(\large\frac{\sec^2\theta-1}{\sec^2\theta}=\cos^2\theta\)
\(cos^2\theta-\sin^2\theta=1-2\sin^2\theta\)
\(\sin^2\theta=\cos^2\theta-1\)

Answer 2

Can you help @rvc?

Answer 3

@phi

Answer 4

http://openstudy.com/study#/updates/553f1223e4b061e2642b5eb6

Answer 5

I'm still not sure..

Answer 6

hmmm

Answer 7

what do u think?

Answer 8

ohmygoodness, I think I have it now.. is it \(\cos^2\theta-\sin^2\theta=1-2\sin^2\theta\)?

Answer 9

that is a double angle formula
cos2x=cos^2-sin^2=1-2sin^x
maybe yeah
but i would like to confirm @perl

Answer 10

because I thought if you would add the \(2\sin^2\theta\) to the other side, it would be \(\cos^\theta+\sin^2\theta=1\), right?

Answer 11

@rvc

Answer 12

oops, I meant \(\cos^2\theta+\sin^2\theta=1\)

Answer 13

we have a double angle formula
\[\Large~ cos2x=cos^2x-sin^2x=1-2sin^2x\]

Answer 14

so is it not correct?..

Answer 15

yeah
i think
but the first option

Answer 16

what about the first option?

Answer 17

if we take sin^2x common fram the RHS then?

Answer 18

both the sin^2x cancels from both the side

Answer 19

tan^2x=sec^2+1

Answer 20

sec^2x*

Answer 21

I'm confused..

Answer 22

wait
\[\Large sin^2\theta.tan^2\theta =\frac{sin^2\theta}{cos^2\theta}+sin^2\theta\]

right?

Answer 23

I don't know.

Answer 24

look at the given link

Answer 25

I think so?

Answer 26

we can also write
\(\huge\color{red}{tanx=\frac{sinx}{cosx}}\)

Answer 27

right?

Answer 28

yes

Answer 29

so got that step?

Answer 30

yes

Answer 31

Oh, I was right, by the way, but I do want you to continue explaining it. :)

Answer 32

Now take sin^2x common from the RHS

Answer 33

what do you get

Answer 34

what do you mean?

Answer 35

did you get this?
or got something else?

Answer 36

how'd you get that? I'm sorry, all of this is very confusing to me

Answer 37

oops
all are squares

Answer 38

okay wait

Answer 39

we have sin^2x . tan^2x=tan^2x + sin^2x

Answer 40

only lets simplify the RHS
\[\huge \tan^2x+\sin^2x\]

Answer 41

@BloomLocke367 u here?

Answer 42

I am, sorry, I was tagged.

Answer 43

what is RHS?