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Trigonometric Identity Problems

Can anyone help me solve the following problems?

sec theta -1/1-cos theta = sec theta

tan (pie/2 - theta) tan theta =1

Thanks
 

James R

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Your first equation is ambiguous. Could you please add appropriate brackets so we know where the fraction is?

For the second one, do you know the identity for tan(A+B)?

Because that would be a good place to start.
 

dextercioby

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For the second, use the definition of the tangent function.

Daniel.
 

HallsofIvy

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Robert R's point is that it is not clear whether you mean sec theta -1/(1-cos theta) = sec theta or (sec theta -1)/(1- cos theta). What is the DEFINITION of sec theta? Then try multiplying both sides of the equation by cos theta.
 
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Krystal i think you meant:

for question 1

[sec(theta) - 1] / [1 - cos(theta)] = sec(theta)

if you did then:
try writing sec(theta) as 1/cos(theta)
then make a common denominator for: 1/cos(theta) -1
and continue from there
 
I have a problem that I cannot get past step one; the problem is verifying the identity:

cos2x(1+cot^2x) = csc^2 - 2 ?????

I know that [1 + cot^2x] = [csc^2x]

Thanks for all the help.
 
nevermind...i got the answer...it was:

(1-2sin^2 x)/sin^2 x = (1-2sin^2 x)/sin^2 x
 
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can anyone help me to solve this problem?
(tanѲ/1-cotѲ)+(cotѲ/1-tanѲ)=1+secѲcscѲ
 

SammyS

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can anyone help me to solve this problem?
(tanѲ/1-cotѲ)+(cotѲ/1-tanѲ)=1+secѲcscѲ
You have posted this on a 4 year old thread.

You may get more help, if you start a new thread and post it there.
 
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trigonometric solver

(tanѲ/1-cotѲ)+(cotѲ/1-tanѲ)=1+secѲcscѲ
 

SammyS

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Without the proper placement of parentheses, or other grouping symbols, what you have written is equivalent to: [itex]\displaystyle \left(\frac{\tan\theta}{1}-\cot\theta\right)+\left(\frac{\cot\theta}{1}-\tan\theta\right)=1+\sec\theta\,\csc\theta\,.[/itex]

Perhaps you meant: (tanθ/{1-cotθ})+(cotθ/(1-tanθ))=1+secθcscθ, which is the same as: [itex]\displaystyle \left(\frac{\tan\theta}{1-\cot\theta}\right)+\left(\frac{\cot\theta}{1-\tan\theta}\right)=1+\sec\theta\,\csc\theta\,.[/itex]
 

SammyS

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To show that [itex]\displaystyle \left(\frac{\tan\theta}{1-\cot\theta}\right)+\left(\frac{\cot\theta}{1-\tan\theta}\right)=1+\sec\theta\,\csc\theta\,[/itex] is an identity, I suggest writing the tangent & cotangent functions in terms of sine & cosine.
 

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