C3 trig!

I can't do these two:

24) Given that cotA=1/4 and cot(A+B)=2, find the value of cotB.
Cot(A+B) = 1-tanAtanB / tanA + tanB = 2, right?? Then what?

25) Given that tan(x + π/3) = 1/2, show that tanx=8-5√3.

Reply 1

steve2005

turtle2

I can't do these two:

24) Given that cotA=1/4 and cot(A+B)=2, find the value of cotB.
Cot(A+B) = 1-tanAtanB / tanA + tanB = 2, right?? Then what?

25) Given that tan(x + ?/3) = 1/2, show that tanx=8-5?3.

Here is first answer ( I think) Error corrected.

Picture 33.png26.8KB

Reply 2

Malik

steve2005

Here is first answer ( I think)

Youve made slight mistake in your formula

tan(A+B)= (tan A+ tan B)/(1-tanAtanB)

Reply 3

steve2005

25) Given that tan(x + ?/3) = 1/2, show that tanx=8-5?3.
Here is the proof

Picture 32.png39.3KB

Reply 4

steve2005

mala2k

Youve made slight mistake in your formula

tan(A+B)= (tan A+ tan B)/(1-tanAtanB)

Thanks.

Reply 5

turtle2

Thanks a lot!

Could you please also help with : given that tan(x-y)=3, express tany in terms of tanx. (I can't get the correct answer!)

Reply 6

Dimez

tanx - tany/(1 + tanxtany) = 3

===> tanx - tany= 3tanxtany

tanx -3= 3tanxtany + tany

tanx - 3= tany(3tanx + 1)

tanx - 3/(3tanx + 1) = tany

Hey Presto!!

Reply 7

Malik

turtle2

Thanks a lot!

Could you please also help with : given that tan(x-y)=3, express tany in terms of tanx. (I can't get the correct answer!)

tan(x-y)=3
(tan x- tan y)/(1+tanxtany)=3
To Make this more simple just say: tanx=x tany=y
(x-y)/(1+xy)=3
x-y=3(1+xy)
x-y=3+3xy
x-3xy=3+y
x(1-3y)=3+y
x=(3+y/(1-3y)

so tanx=(3+tany)/(1-3tany)

EDIT: :redface: