Can Trigonometric Identities Be Proven Using Different Methods?

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Homework Help Overview

The discussion revolves around proving trigonometric identities, specifically the identity involving tangent and cotangent functions. Participants explore different methods of proof and the reasoning behind their approaches.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to find alternative methods for proving a trigonometric identity after initially solving it using sine and cosine. Some participants question the necessity of seeking another method and discuss the equivalence of expressions. Others suggest manipulating the identity using different trigonometric relationships and common denominators.

Discussion Status

The discussion is active, with participants providing insights into alternative approaches and questioning the original poster's reasoning. Some guidance has been offered regarding the manipulation of trigonometric functions, but there is no explicit consensus on the best method to prove the identity.

Contextual Notes

Participants also introduce a new question regarding the slopes of perpendicular lines, indicating a shift in focus within the discussion. There is an exploration of geometric interpretations and trigonometric relationships related to angles between lines.

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 [itex]\displaystyle \frac{\tan^2(\phi)}{\tan(\phi)-1}+\frac{\cot^2(\phi)}{\cot(\phi)-1}=1+\sec(\phi)\csc(\phi)\ ,[/itex]

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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