# precalc h

posted by on .

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

• precalc h - ,

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
when typing on the computer i can not tell what you mean for numerators or denominators.

• error fx - ,

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

• precalc h - ,

The parenthesis were for the arguments.. My question was how do I verify this indentity

• precalc h - ,

Theta = argument..

• precalc h - ,

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 - ,

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)

• precalc h - ,

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.

• precalc h - ,

I don't understand what you did?

• precalc h - ,

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

• precalc h - ,

• precalc h - ,

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 - ,

Only work on the left side of your equal sign

Your LCD on the left is
(1-cos 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 (1-cos t)times (1-cos t) you get (1-cos t)^2

• precalc h - ,

1+1/cos = tan^2/sec-1

I will guess (parentheses missing again) you mean

1+1/cos = tan^2/(sec-1 )

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(1-cos)

(cos + 1)/cos = (1-cos^2)/cos(1-cos)
but (1-cos^2)= (1+cos)(1-cos)

(cos + 1)/cos = (1+cos)(1-cos)/cos(1-cos)

(cos + 1)/cos = (1+cos)/cos

• precalc h - ,

dank memes