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# Trig identity.Help watch

1. Need help in these two questions
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2. First part of first question from double angle identities, then it's a case of rearranging and factorising until your left with sinA/CosA = tanA
3. Second question square both sides to start and then rewrite the cos2x term and start working towards something that could resemble tanx
4. (Original post by mathcoachni)

First part of first question from double angle identities, then it's a case of rearranging and factorising until your left with sinA/CosA = tanA
I did this part so far already and with one more step which is factorising Cos^2(A)-Sin^2(A) but got stuck after that
5. (Original post by lil_jack)
I did this part so far already and with one more step which is factorising Cos^2(A)-Sin^2(A) but got stuck after
Remember what your trying to make the LHS look like, you want sin/cos, sometimes it just takes you to go through the different options.
6. (Original post by lil_jack)
I did this part so far already and with one more step which is factorising Cos^2(A)-Sin^2(A) but got stuck after that
remember that

this is done by using

now in the denominator, the -1 and +1 cancel out each other, so you are left with:

Factorise this now and you should be able to get the answer
7. Thanks alot mate

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