precalc h
posted by sam .
sin(theta)/1cos(theta) + 1cos(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/sec1

precalc h 
Damon
maybe you mean
sin t/(1cos t) + (1cos t)/sin t
sin^2 t (1cos t) (1cos t)^2
 + 
(1cos t)(sin t) ______ (1cos t)(sin t)
numerator only for a while
sin^2 tsin^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

(1cos 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. 
error fx 
Damon
sin t/(1cos t) + (1cos t)/sin t
sin^2 t ______________ (1cos t)^2
 + 
(1cos t)(sin t) ______ (1cos t)(sin t)
sin^2 t + 1 2 cos t +cos^2 t

(1cos t)(sin t)
2  2 cos t

(1cos t) sin t
= 2/sin t
which is csc t 
precalc h 
sam
The parenthesis were for the arguments.. My question was how do I verify this indentity

precalc h 
sam
Theta = argument..

precalc h 
Damon
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 
precalc h 
Damon
Here is how you should have written it
(use t for theta)
sin(theta)/ { 1cos(theta)} + {1cos(theta)}/sin(theta) = 2 csc(theta) 
precalc h 
Damon
The critical point is that afteter you get it all over the LCD (sin t)(1cos t)
you see that you have sin^2 t+ cos^2 t which is one in the numerator. 
precalc h 
sam
I don't understand what you did?

precalc h 
sam
How would x'ing the right side of the plus side by sin give you (1cos)^2

precalc h 
sam
OH nevermind about that .

precalc h 
sam
okay I get it now, thanks so much. I have another one at the bottom of this question.. could you help me with it? But THANKYOU SO MUCHCHCH! I get it. :)

precalc h 
Damon
Only work on the left side of your equal sign
Your LCD on the left is
(1cos t)(sin t)
to get that on the bottom of the first term you multiply top and bottom by sin t
to get it on the bottom of the second term you multiply top and bottom by (1 cos t)
when you multiply (1cos t)times (1cos t) you get (1cos t)^2 
precalc h 
Damon
1+1/cos = tan^2/sec1
I will guess (parentheses missing again) you mean
1+1/cos = tan^2/(sec1 )
use sin and cos everywhere
1+1/cos = (sin/cos)^2/([1/cos]1 )
multiply top and bottom of right by cos^2
1 + 1/cos = sin^2/(cos  cos^2)
(cos + 1)/cos = sin^2/cos(1cos)
(cos + 1)/cos = (1cos^2)/cos(1cos)
but (1cos^2)= (1+cos)(1cos)
(cos + 1)/cos = (1+cos)(1cos)/cos(1cos)
(cos + 1)/cos = (1+cos)/cos 
precalc h 
lolwut
dank memes
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