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A level Trig Help!

Hello, can someone help me with these two questions please:

(a) Prove: cotx - tanx = 2cot2x (DONE)

(b) Prove 2tanx / (1 + tan(^2)x = sin2x (DONE)

Thank you so much in advance! :smile:

Edit: and this question too please:

Prove (cosecA + cotA)^2 = 1 + cosA / 1 - cosA
(edited 9 years ago)
Original post by chococup123
Hello, can someone help me with these two questions please:

(a) Prove: cotx - tanx = 2cot2x

(b) Prove 2tanx / (1 + tan(^2)x = sin2x

Thank you so much in advance! :smile:

Edit: and this question too please:

Prove (cosecA + cotA)^2 = 1 + cosA / 1 - cosA

I'd probably start with the RHS for (a).
And the LHS for (b). Where tanx\text tanx is involved, I might be tempted to express it interms of sinx\text sinx and cosx\text cosx and when it's squared, I'd think about what links tan2x\text tan^2x and 1\text 1.
(edited 9 years ago)
Original post by keromedic
I'd probably start with the RHS for (a).
And the LHS for (b). Where tanx\text tanx is involved, I might be tempted to express it interms of sinx\text sinx and cosx\text cosx and when it's squared, I'd think about what links tan2x\text tan^2x and 1\text 1.


Oh but how would you do part (a) ?
Original post by chococup123
Oh but how would you do part (a) ?

RHS=2cot2x=2cos2xsin2x=...RHS=2cot2x=\dfrac{2cos2x}{sin2x}=...
Original post by keromedic
RHS=2cot2x=2cos2xsin2x=...RHS=2cot2x=\dfrac{2cos2x}{sin2x}=...


then after that, do you do:

2 cos(^2)x - 2sin(^2)x / 2sinxcosx
Original post by chococup123
then after that, do you do:

(2cos2x - 2sin2x )/(2sinxcosx)

Yup :smile:. Also, please use brackets to prevent ambiguity.
(edited 9 years ago)
Original post by keromedic
Yup :smile:. Also, please use brackets to prevent ambiguity.


Yeah, thank you! :smile: where do I go from here? Sorry.
Original post by chococup123
Yeah, thank you! :smile: where do I go from here? Sorry.

Do you really have no idea? Try and compare what you have to the other side. Hopefully, it should jump out at you.
It'll be helpful for you to remember that abc+d=ac+dbc+d\dfrac{a-b}{c+d}=\dfrac{a}{c+d}-\dfrac{b}{c+d}.
Original post by keromedic
Do you really have no idea? Try and compare what you have to the other side. Hopefully, it should jump out at you.
It'll be helpful for you to remember that abc+d=ac+dbc+d\dfrac{a-b}{c+d}=\dfrac{a}{c+d}-\dfrac{b}{c+d}.


Oh yeah, sorry! that was really stupid of me! haha thank you so much for your help though. Also would you help me on part c? (first post, edit)
Is part (c) (cosecA+cotA)2=1+cosA1cosA (cosecA + cotA)^2 = \dfrac{1 + cosA}{ 1 - cosA}?
If so, I don't think it matters which side you start from. Both are a little tricky. Might want to do a bit of both.
Original post by keromedic
Is part (c) (cosecA+cotA)2=1+cosA1cosA (cosecA + cotA)^2 = \dfrac{1 + cosA}{ 1 - cosA}?
If so, I don't think it matters which side you start from. Both are a little tricky. Might want to do a bit of both.


Yeah, that's part (c) And oh. Would I annoy you if I asked you to start me off? I am so sorry for being such a pain and thank you for your kindness

This is what I've done:

(1/sinA + 1/tanA)^2

= 1/sin(^2)A + 1/2SinAtanA + 1/tan(^2)A
(edited 9 years ago)
Original post by chococup123


(b) Prove 2tanx / (1 + tan(^2)x = sin2x


A general rule of thumb - turn tan(x) into sin(x) and cos(x)

2sinxcosx1+sin2xcos2x\dfrac{\frac{2\sin x}{\cos x}}{1+\frac{\sin ^2x}{\cos ^2x}}

Also remove fractions in fractions by multiplying numerator and denominator (in this case multiply by cos2x\cos ^2x


Also think ahead to what the answer will need to look like
Original post by chococup123
Yeah, that's part (c) And oh. Would I annoy you if I asked you to start me off? I am so sorry for being such a pain and thank you for your kindness

You could begin saying that RHS=1+cosA1cosA=1+cosA1cosA×1+cosA1+cosARHS=\dfrac{1+cosA}{1-cosA} = \dfrac{1+cosA}{1-cosA} \times \dfrac{1+cosA}{1+cosA}.

Edit: You're doing the LHS and ten of them is helping you :smile:.
(edited 9 years ago)
Original post by TenOfThem
A general rule of thumb - turn tan(x) into sin(x) and cos(x)

2sinxcosx1+sin2xcos2x\dfrac{\frac{2\sin x}{\cos x}}{1+\frac{\sin ^2x}{\cos ^2x}}

Also remove fractions in fractions by multiplying numerator and denominator (in this case multiply by cos2x\cos ^2x


Also think ahead to what the answer will need to look like


Oh thank you! I've worked out (b)
Original post by chococup123
Yeah, that's part (c) And oh. Would I annoy you if I asked you to start me off? I am so sorry for being such a pain and thank you for your kindness

This is what I've done:

(1/sinA + 1/tanA)^2

= 1/sin(^2)A + 1/2SinAtanA + 1/tan(^2)A


urg - learn LaTeX :biggrin:

AS I have said sin(x) and cos(x)

(1sinA+cosAsinA)2=1sin2A+2cosAsin2A+cos2Asin2A(\dfrac{1}{\sin A} + \dfrac{\cos A}{\sin A})^2 = \dfrac{1}{\sin ^2A} + \dfrac{2\cos A}{\sin^2 A} + \dfrac{\cos^2 A}{\sin ^2A}


Add the fractions

Turn everything into Cos A - because that is what the answer as in it
(edited 9 years ago)
Original post by TenOfThem
urg - learn LaTeX :biggrin:

AS I have said sin(x) and cos(x)

(1sinA+cosAsinA)2=1sin2A+2cosAsin2A+cos2Asin2A(\dfrac{1}{\sin A} + \dfrac{\cos A}{\sin A})^2 = \dfrac{1}{\sin ^2A} + \dfrac{2\cos A}{\sin^2 A} + \dfrac{\cos^2 A}{\sin ^2A}


Add the fractions

Turn everything into Cos A - because that is what the answer as in it


Oh, yeah thank you!
Thank you all! I've done all three questions :smile: :smile:

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