# Trigonometric Identity Problems

#### krystalanderson

Can anyone help me solve the following problems?

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

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

Thanks

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#### James R

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Gold Member
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

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

Daniel.

#### HallsofIvy

Homework Helper
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.

#### sara_87

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.

(1-2sin^2 x)/sin^2 x = (1-2sin^2 x)/sin^2 x

#### bomb

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.

#### bomb

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: $\displaystyle \left(\frac{\tan\theta}{1}-\cot\theta\right)+\left(\frac{\cot\theta}{1}-\tan\theta\right)=1+\sec\theta\,\csc\theta\,.$

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

#### SammyS

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