Question

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Which of the following are identities? More than one can be the answer.

Answer 1

Well when you get both sides to equal eachother that is an ID

Answer 2

If you want to be able to answer these kind of questions, there is one thing you need to do first:

Learn by heart all the basic identities.

I mean like:

sin²x+cos²x=1
sin2x=2sinxcosx
cos2x=1-2sin²x=2cos²x-1=cos²x-sin²x

sin(x+y)=sinxcosy+cosxsiny
cos(x+y)=cosxcosy-sinxsiny

And there is still more...

Answer 3

OK, because I want to help you and not scare you ;), let's try the first one:

You can see sin(x+y) on the right hand side. There is an identity for this:

\[\frac{ \sin x \cos y+\cos x \sin y }{ \sin x \sin y }\]

What to do next? Well, on the LHS, there is no sin in sight, only tangents there.
We have to find a way to get there!
Now you know that tanx = sinx/cosx, so if we split the fraction we get:\[\frac{ \sin x \cos y }{ \sin x \sin y }+\frac{ \cos x \sin y }{ \sin x \sin y }=\frac{ \cos y }{ \sin y }+ \frac{ \cos x }{ \sin x }=\frac{1}{tan y}+\frac{1}{tan x}\]Ok, this is not yet the LHS, but we're getting closer!

Answer 4

In fact, if we write it as one fraction we get:\[\frac{ \tan x +\tan y }{ \tan x \tan y }\]What we have here now, is a nice identity: \[\frac{ \tan x +\tan y }{ \tan x \tan y }=\frac{ \sin(x+y) }{ \sin x \sin y }\]BUT: it is not "identity" A...

You see: a lot of work, just to write off A.

3 other options to go...

Answer 5

I got B and D