Homework Help: Trigonometric Identity

1. May 5, 2012

physics kiddy

1. The problem statement, all variables and given/known data

Prove that:
tan^2∅/tan∅ - 1 + cot^2∅/cot∅ - 1 = 1 + sec∅cosec∅

2. Relevant equations

3. The attempt at a solution

I have solved the question taking tan∅ = sin∅/cos∅.
But I want to solve it some other way.

2. May 5, 2012

micromass

What other way??
Why aren't you happy with your solution?

3. May 5, 2012

SammyS

Staff Emeritus
What you wrote for the left hand side is literally (tan^2∅/tan∅) - 1 + (cot^2∅/cot∅) - 1, which is equivalent to tan∅ + cot∅ - 2 .

Assuming that you meant $\displaystyle \frac{\tan^2(\phi)}{\tan(\phi)-1}+\frac{\cot^2(\phi)}{\cot(\phi)-1}=1+\sec(\phi)\csc(\phi)\ ,$

yes there is another way. --- it's no better, but looks interesting enough. Even with it, eventually you will change tan to sin/cos or perhaps tan to sec/csc.

Change cot(ϕ) to 1/tan(ϕ) . Then multiply the numerator & denominator of the second fraction by -tan(ϕ) --- that will give you a common denominator. You can then get a difference of cubes in the numerator ...

4. May 5, 2012

physics kiddy

Thanks, I got the answer. But I have got one more question:

How to prove that slopes of perpendicular lines on graph paper have a product equal to -1 ?

5. May 6, 2012

Infinitum

What is the angle(acute) between two lines of slopes say, m1 and m2?
When will they become perpendicular then?

Last edited: May 6, 2012
6. May 6, 2012

No idea !!!

7. May 6, 2012

Infinitum

Okay hmm, try drawing out two lines with a general angle θ between them. Say the angle the first line makes with the positive x axis is A and the second line makes an angle B, now try finding a trignometrical relation between θ, A and B. (Hint: use the property of external angles)

8. May 6, 2012

physics kiddy

I have attached a pic. Tell me if it is like that.

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9. May 6, 2012

Infinitum

The x axis is not necessarily where the two lines meet. So you can draw them cutting the x axis at different points, and such that they intersect somewhere arbitarily on the xy plane, for the sake of a more general result.