Can someone help me with C4 Trigonometry double angles question?

"Prove the given identity:sin2x = 2tanx/(1 + tan^2x)"
Could someone show me how they would do this? Before anyone tells me to look at a mark scheme, this isn't an exam question, this is a revision question from a book, and no method is given. I've already tried it but I'm not sure if my method is correct. I've put my attempt at it below - is it right?
My answer:
sin2x
= 2sinxcosx
= 2(tanxcosxcosx) <-- (can you even change sinx to tanxcosx, using the identity tanx = sinx/cosx ???)
= 2tanxcos^2x
= 2tanx(1-sin^2x)
=2tanx - 2tanxsin^2x
= 2tanx -2tanx . 1/cosec^2x
= 2tanx - 2tanx/cosec^2x
= (2tanxcosec^2x - 2tanx)/cosec^2x
= (2tanx (cosec^2 - 1))/cosec^2x
= (2tanxcot^2x)/(1+cot^2x)
= (2tanx/tan^2x)/(1+ 1/tan^2x)
= 2tanx/ (tan^2x(1+1/tan^2x))
= 2tanx/(tan^2x +1)
= 2tanx/ (1+tan^2x) as required.

Sorry it's so long and thanks in advance!

So basically you should start with the right hand side. the denominator is one of the identities 1+tan^2x = sec^2x
so if you plug that in you can cancel out using tanx = sinx/cosx

Idk if im making sense but I used it and got the answer

...

"Prove the given identity:sin2x = 2tanx/(1 + tan^2x)"
Could someone show me how they would do this? Before anyone tells me to look at a mark scheme, this isn't an exam question, this is a revision question from a book, and no method is given. I've already tried it but I'm not sure if my method is correct. I've put my attempt at it below - is it right?
My answer:
sin2x
= 2sinxcosx
= 2(tanxcosxcosx) <-- (can you even change sinx to tanxcosx, using the identity tanx = sinx/cosx ???)
= 2tanxcos^2x
= 2tanx(1-sin^2x)
=2tanx - 2tanxsin^2x
= 2tanx -2tanx . 1/cosec^2x
= 2tanx - 2tanx/cosec^2x
= (2tanxcosec^2x - 2tanx)/cosec^2x
= (2tanx (cosec^2 - 1))/cosec^2x
= (2tanxcot^2x)/(1+cot^2x)
= (2tanx/tan^2x)/(1+ 1/tan^2x)
= 2tanx/ (tan^2x(1+1/tan^2x))
= 2tanx/(tan^2x +1)
= 2tanx/ (1+tan^2x) as required.

Sorry it's so long and thanks in advance!

haha thanks! i should've just done that

Your proof is a little long compared to the intended method, but it definitely works (haven't checked it thoroughly though).

One important lesson you can take away from this: if you're asked to prove an identity, you may start from either side and get to the other. You don't need to go from left to right. Starting with the RHS and ending up with the LHS is just as valid a proof.

You could have simplified it a lot more by knowing that $cos^2x \equiv \dfrac{1}{sec^2x} \equiv \dfrac{1}{1+tan^2x}$.

You got there but it took a lot of lines.

I would have done it differently after the bolded lines.

"Prove the given identity:sin2x = 2tanx/(1 + tan^2x)"
Could someone show me how they would do this? Before anyone tells me to look at a mark scheme, this isn't an exam question, this is a revision question from a book, and no method is given. I've already tried it but I'm not sure if my method is correct. I've put my attempt at it below - is it right?
My answer:
sin2x
= 2sinxcosx
= 2(tanxcosxcosx) <-- (can you even change sinx to tanxcosx, using the identity tanx = sinx/cosx ???)
= 2tanxcos^2x
= 2tanx(1-sin^2x)
=2tanx - 2tanxsin^2x
= 2tanx -2tanx . 1/cosec^2x
= 2tanx - 2tanx/cosec^2x
= (2tanxcosec^2x - 2tanx)/cosec^2x
= (2tanx (cosec^2x - 1))/cosec^2x
= (2tanxcot^2x)/(1+cot^2x)
= (2tanx/tan^2x)/(1+ 1/tan^2x)
= 2tanx/ (tan^2x(1+1/tan^2x))
= 2tanx/(tan^2x +1)
= 2tanx/ (1+tan^2x) as required.

Sorry it's so long and thanks in advance!

you missed out an x... :spank: but otherwise it's watertight.