# Trigonometric Identity Problems

1. Jan 4, 2007

### 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

2. Jan 5, 2007

### James R

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.

3. Jan 5, 2007

### dextercioby

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

Daniel.

4. Jan 5, 2007

### HallsofIvy

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.

5. Jan 5, 2007

### 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

6. Jul 16, 2007

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.

7. Jul 16, 2007

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

8. Jul 25, 2011

### bomb

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

9. Jul 25, 2011

### SammyS

Staff Emeritus
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.

10. Jul 26, 2011

### bomb

trigonometric solver

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

11. Jul 26, 2011

### SammyS

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

12. Jul 26, 2011

### 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.