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

 

[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.

 

[dextercioby]

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

Daniel.

 

[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.

 

[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

 

[dadagladinian]

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.

 

[dadagladinian]

nevermind...i got the answer...it was:

(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]

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]

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]

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.