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May 30, 2016
Posted by **sam** on Sunday, December 9, 2012 at 8:17pm.

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

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**Damon**, Sunday, December 9, 2012 at 8:36pmmaybe 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. - error fx -
**Damon**, Sunday, December 9, 2012 at 8:48pmsin 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 -
**sam**, Sunday, December 9, 2012 at 8:51pmThe parenthesis were for the arguments.. My question was how do I verify this indentity

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**sam**, Sunday, December 9, 2012 at 8:54pmTheta = argument..

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**Damon**, Sunday, December 9, 2012 at 9:09pmI 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**, Sunday, December 9, 2012 at 9:11pmHere 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 -
**Damon**, Sunday, December 9, 2012 at 9:13pmThe 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 -
**sam**, Sunday, December 9, 2012 at 9:17pmI don't understand what you did?

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**sam**, Sunday, December 9, 2012 at 9:20pmHow would x'ing the right side of the plus side by sin give you (1-cos)^2

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**sam**, Sunday, December 9, 2012 at 9:21pmOH nevermind about that .

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**sam**, Sunday, December 9, 2012 at 9:24pmokay 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. :--)

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**Damon**, Sunday, December 9, 2012 at 9:27pmOnly 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 -
**Damon**, Sunday, December 9, 2012 at 9:35pm1+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