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Quick C3 Maths Question

Hello,

I have to simply this expression, but I am stuck on where to start.

I'm forgetting something here, and I can't put my finger on it.

So, this is the expression, : 1/tan(x)+cot(x), i need this in it's simplest form.

Any help is appreciated.
Reply 1
Avatar for Slumpy
Slumpy
16
Are you asking 1/(tanx+cotx) or (1/tanx)+cotx?
Reply 2
Avatar for ekudamram
ekudamram
what is the relation between tanx and cotx?
Reply 3
Avatar for C94
C94
OP
1
Original post by Slumpy
Are you asking 1/(tanx+cotx) or (1/tanx)+cotx?


1/(tanx+cotx)
Reply 4
Avatar for Slumpy
Slumpy
16
Original post by C94
1/(tanx+cotx)


Rewrite it with the definitions of tan and cot. Should be fairly clear where to go from there.
Reply 5
Avatar for C94
C94
OP
1
Original post by ekudamram
what is the relation between tanx and cotx?


Cotx is the reciprocal of tanx.
(edited 11 years ago)
Reply 6
Avatar for ekudamram
ekudamram
Original post by C94
Cotx is the reciprocal of tanx.


So why not rewrite them both in terms of one (e.g. rewrite cotx = 1/tanx)
Reply 7
Avatar for C94
C94
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1
Original post by Slumpy
Rewrite it with the definitions of tan and cot. Should be fairly clear where to go from there.


I did do that, and I got sinxcosx/sinx+cosx. :/
Reply 8
Avatar for C94
C94
OP
1
Can someone give me the full answer?
Reply 9
Avatar for ekudamram
ekudamram
What was the expression initially?
Reply 10
Avatar for C94
C94
OP
1
1/tanx+cotx, i need it in simple form.
Reply 11
Avatar for Slumpy
Slumpy
16
Original post by C94
I did do that, and I got sinxcosx/sinx+cosx. :/


You've made an error in the denominator.
Reply 12
Avatar for ekudamram
ekudamram
Forget the ^-1 part of the equation for now, lets focus on tanx+cotx

tanx+cotx = sinx/cosx + cosx/sinx (we need common denominator)

= sin^2x/cosxsinx + cos^2x/cosxsinx (sin^2x + cos^2x = 1)

= 1/cosxsinx

but remember it was originally 1/tanx+cotx therefore it is the inverse of this

therefore 1/(1/cosxsinx) = cosxsinx
Reply 13
Avatar for C94
C94
OP
1
Original post by ekudamram
Forget the ^-1 part of the equation for now, lets focus on tanx+cotx

tanx+cotx = sinx/cosx + cosx/sinx (we need common denominator)

= sin^2x/cosxsinx + cos^2x/cosxsinx (sin^2x + cos^2x = 1)

= 1/cosxsinx

but remember it was originally 1/tanx+cotx therefore it is the inverse of this

therefore 1/(1/cosxsinx) = cosxsinx


Thanks for that,, :smile:
Reply 14
Avatar for ekudamram
ekudamram
No worries

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