T method help (1 Viewer)

[Drewk]

http://snag.gy/QOLLm.jpg
Please solve i am getting them wrong

 

[Kurosaki]

Tan(theta) is =(2t)/(1-t^2)

Cos(theta)= (1-t^2)/(1+t^2)
Sec is the reciprocal of cos
Then you just add them.

 

[Drewk]

Tan(theta) is =(2t)/(1-t^2)

Cos(theta)= (1-t^2)/(1+t^2)
Sec is the reciprocal of cos

Can u plz show ur working i already knew all that but still got it wrong

 

[Kurosaki]

Can u plz show ur working i already knew all that but still got it wrong

Sorry dude it's a little hard typing on an iPod and I have low battery

 

[Kurosaki]

Ok given the values for tan and sec I gave you
1 + tan(theta)= 1 + 2t/(1-t^2). Adding them, you get (1-t^2+2t)/(1-t^2)
Then
Tan(theta) + sec(theta)= (1-t^2+2t)/(1-t^2) + (1+t^2)/(1-t^2)
U get to (2t + 2)/(1-t^2)

Divide by (1+t) u get 2/(1-t) ?

 

[Drewk]

Look at answers here according to character
http://snag.gy/JckJs.jpg

 

[Drewk]

Ok given the values for tan and sec I gave you
1 + tan(theta)= 1 + 2t/(1-t^2). Adding them, you get (1-t^2+2t)/(1-t^2)
Then
Tan(theta) + sec(theta)= (1-t^2+2t)/(1-t^2) + (1+t^2)/(1-t^2)
U get to (2t + 2)/(1-t^2)

Divide by (1+t) u get 2/(1-t) ?

I still dont get part i

 

[Kurosaki]

According to my working my answers are right...let me work on it again

 

[Drewk]

According to my working my answers are right...let me work on it again

Tan(theta) + sec(theta)= (1-t^2+2t)/(1-t^2) + (1+t^2)/(1-t^2)

in ur line here ^ in 1st bracket, where did u get 1-t^2 from ?? Tan(theta) is =(2t)/(1-t^2)

 

[Kurosaki]

Tan(theta) + sec(theta)= (1-t^2+2t)/(1-t^2) + (1+t^2)/(1-t^2)

in ur line here ^ in 1st bracket, where did u get 1-t^2 from ?? Tan(theta) is =(2t)/(1-t^2)

Lol whoops I used the result from the first question. Lol

Ok then the working would be
2t/(1-t^2) + (1+t^2)/(1 -t^2)
You get like
(t + 1)^2/(1-t^2)
Divide by (t+1)
U get (t+1)/(1-t)

 

[Kurosaki]

Sorry for the long wait I just type really slow on an iPod

 

[Drewk]

Lol whoops I used the result from the first question. Lol

Ok then the working would be
2t/(1-t^2) + (1+t^2)/(1 -t^2)
You get like
(t + 1)^2/(1-t^2)
Divide by (t+1)
U get (t+1)/(1-t)

Why Divide by (t+1)

 

[Kurosaki]

Difference of two squares

 

[Drewk]

Difference of two squares

The division seems a Bit random, but ty

 

[Kurosaki]

No worries bro. Kurosaki's the name and maths is my game

 

[Aysce]

The division seems a Bit random, but ty

Because you saw (t+1) on both the numerator and denominator in which they must cancel.

 

[Kurosaki]

So anything else dude? I finally moved to a computer so my replying speed has increased exponentially

 

[yasminee96]

this is my working, a tad messy coz hand is injured sorry :O

http://snag.gy/BxyxA.jpg




EDIT: pic was taken on a webcam, terrible quality; sorry !

 

[Sy123]

this is my working, a tad messy coz hand is injured sorry :O

http://snag.gy/BxyxA.jpg


EDIT: pic was taken on a webcam, terrible quality; sorry !

There is a mistake in the 5th line of your working for the first part.


=========================


Anyway, for the question:





That is the furthest simplification without going back to the start, unless you want to do something ridiculous like:

 

[yasminee96]

There is a mistake in the 5th line of your working for the first part.


[/tex]

you're right, woopsy; thanks.

well the required answer is there aha