trig equations help (1 Viewer)

DeeDeeChan · Mar 30, 2009

i just finished a maths exam and had a whole lot of trouble with a certain trig equation:

3/sec(THETA)-1 - 3/sec(THETA) + 1 = 6cot^2(THETA)

it could be that fact i didnt study it in enough depth but i would still like to know how to do this equation.

thankyou :wave:

Drongoski · Mar 30, 2009

Is that an identity to be proved ?? Then you have stated it not quite correctly.

$LHS \ \ =\ \ \frac{3}{sec\ \theta - 1}-\frac{3}{sec\ \theta +1}\ \ =\ \ \frac{3(sec\ \theta \ +1)-3(sec\ \theta \ -\ 1)}{sec^{2}\ \theta - 1}\\ \\ \ \ \ =\ \ \frac{6}{tan^{2 }\theta }\ \ =\ \ 6\ cot^{2}\ \theta\ \ =\ \ RHS\\ \\ \\ QED$

DeeDeeChan · Mar 31, 2009

oh ok, yeah we were meant to.
makes more sense now, i guess i made a mistake between the sec and tan~~

trig equations help (1 Viewer)

DeeDeeChan

New Member

Drongoski

Well-Known Member

DeeDeeChan

New Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)