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Prove that tan(x) + cot(x)= 2cosec2(x)
Original post by Honere
Prove that tan(x) + cot(x)= 2cosec2(x)


To start off, try to express tan(x) and cot(x) in terms of cos(x) and sin(x)
Reply 2
Original post by ManLike007
To start off, try to express tan(x) and cot(x) in terms of cos(x) and sin(x)


I got sinx/cosx+cosx/sinx
Original post by Honere
I got sinx/cosx+cosx/sinx


That's right, now can you simplify it into one fraction? The numerator should now be familiar, does it ring any bell?
Original post by Honere
Prove that tan(x) + cot(x)= 2cosec2(x)


tan x + cot x = [(sinx)/(cosx) + (cosx)/(sinx)]
therefore tan x + cot x = [(sinx)^2 + (cosx)^2)/(cosx)(sinx)]

Use (sinx)^2 + (cosx)^2 = 1
to give [1/(cosx)(sinx)]

Then use sin2x = 2(sinx)(cosx)

to equal 2/sin2x

Finally, use identity: 1/sinx = cosecx

To end up with 2cosec2x

Therefore tan(x) + cot(x)= 2cosec2(x)
Reply 5
Original post by ManLike007
That's right, now can you simplify it into one fraction? The numerator should now be familiar, does it ring any bell?


I got sin^2x+cos^2x/cosx*sinx
Original post by Honere
I got sin^2x+cos^2x/cosx*sinx


Basic identity you should always remember is sin2x+cos2x=1\sin^{2}x + \cos^{2}x =1 and use this for your numerator.

If I told you 2sinxcosx=sin2x, are you able to use this and simplify the denominator?

Once you do this, the rest should be easy now.
(edited 6 years ago)
Reply 7
Original post by Den987
tan x + cot x = [(sinx)/(cosx) + (cosx)/(sinx)]
therefore tan x + cot x = [(sinx)^2 + (cosx)^2)/(cosx)(sinx)]

Use (sinx)^2 + (cosx)^2 = 1
to give [1/(cosx)(sinx)]


Then use sin2x = 2(sinx)(cosx)

to equal 2/sin2x

Finally, use identity: 1/sinx = cosecx

To end up with 2cosec2x

Therefore tan(x) + cot(x)= 2cosec2(x)


Where did u get 2/sin2x
Original post by Honere
Where did u get 2/sin2x


You need to use the identity 2sinxcosx=sin2x (Hint: make sinxcosx the subject), substitute this into the denominator then you'll get 2sin2x\frac{2}{\sin2x}
Reply 9
Original post by ManLike007
You need to use the identity 2sinxcosx=sin2x (Hint: make sinxcosx the subject), substitute this into the denominator then you'll get 2sin2x\frac{2}{\sin2x}

Thanks I got it

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