Can Trigonometric Identities Be Proven Using Different Methods?

AI Thread Summary
The discussion revolves around proving the trigonometric identity tan²∅/(tan∅ - 1) + cot²∅/(cot∅ - 1) = 1 + sec∅cosec∅. One participant initially solved the problem using tan∅ = sin∅/cos∅ but sought alternative methods. Another contributor clarified the expression and suggested a method involving changing cot(ϕ) to 1/tan(ϕ) and manipulating the fractions for a common denominator. The conversation then shifted to proving that the slopes of perpendicular lines have a product of -1, with hints provided for finding the angle between two lines. The thread highlights the exploration of different approaches to trigonometric proofs and related geometric concepts.
physics kiddy
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Homework Statement



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

Homework Equations





The Attempt at a Solution



I have solved the question taking tan∅ = sin∅/cos∅.
But I want to solve it some other way.
 
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What other way??
Why aren't you happy with your solution?
 
physics kiddy said:

Homework Statement



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

Homework Equations



The Attempt at a Solution



I have solved the question taking tan∅ = sin∅/cos∅.
But I want to solve it some other way.
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 ...
 
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 ?
 
physics kiddy said:
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 ?
What is the angle(acute) between two lines of slopes say, m1 and m2?
When will they become perpendicular then?
 
Last edited:
No idea !
 
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)
 
I have attached a pic. Tell me if it is like that.
 

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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.
 
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