sin(theta)/1-cos(theta) + 1-cos(theta)/sin(theta) = 2csc(theta)
That question makes absolutely no sense.. could someone help me? or lead me in the direction to figuring it out?
and..
(Beside the trig functions is theta)
1+1/cos = tan^2/sec-1
maybe you mean
sin t/(1-cos t) + (1-cos t)/sin t
sin^2 t (1-cos t) (1-cos t)^2
----------------- + ----------------
(1-cos t)(sin t) ______ (1-cos t)(sin t)
numerator only for a while
sin^2 t-sin^2 t cos t +1 -2 cos t+cos^2 t
but sin^2 + cos^2 = 1
2 -sin^2 t cos t - 2 cos t
put denominator back
2 -sin^2 t cos t - 2 cos t
--------------------------
(1-cos t)(sin t)
I think I misunderstood your lack of parentheses but I think you can get the idea
BE CAREFUL ABOUT PARENTHESES!!
when typing on the computer i can not tell what you mean for numerators or denominators.
sin t/(1-cos t) + (1-cos t)/sin t
sin^2 t ______________ (1-cos t)^2
----------------- + ----------------
(1-cos t)(sin t) ______ (1-cos t)(sin t)
sin^2 t + 1 -2 cos t +cos^2 t
--------------------------------
(1-cos t)(sin t)
2 - 2 cos t
--------------
(1-cos t) sin t
= 2/sin t
which is csc t
The parenthesis were for the arguments.. My question was how do I verify this indentity
Theta = argument..
I did verify it.
I only did the left side, starting with what you gave me and ending with
2/sin t
which is 2 csc t
which is the right side
Here is how you should have written it
(use t for theta)
sin(theta)/ { 1-cos(theta)} + {1-cos(theta)}/sin(theta) = 2 csc(theta)
The critical point is that afteter you get it all over the LCD (sin t)(1-cos t)
you see that you have sin^2 t+ cos^2 t which is one in the numerator.