Proving 1 + tan^2X = 1 / cos^2X for 0 < X < 90 in a Right Triangle

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Homework Help Overview

The discussion revolves around proving the trigonometric identity 1 + tan²X = 1 / cos²X within the context of a right triangle, specifically for angles X between 0 and 90 degrees. The triangle is defined with sides labeled as the hypotenuse (c), adjacent side (a), and opposite side (b).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the definitions of tangent and cosine in relation to the sides of the triangle. There is an attempt to substitute these definitions into the original equation. One participant suggests using the Pythagorean theorem to express c² in terms of a and b.

Discussion Status

The discussion is active with participants providing hints and guidance. One participant expresses a newfound clarity regarding the problem, indicating that the approach is becoming clearer, though no consensus or final solution has been reached.

Contextual Notes

Participants are working within the constraints of the problem statement and the properties of right triangles, specifically focusing on the relationships between the sides and angles without additional information or assumptions.

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Homework Statement



Use the given triangle to prove that for 0 < X <90, 1 + tan^2X = 1 / cos^2X

(the given triangle is right angled with angle X marked. The hypotenuse is labeled c, adjacent angle is labeled a and the opposite angle is labeled b)

The Attempt at a Solution



I have no idea where to begin,
tanX = b/a
cosX = a/c
 
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Hi crazy_v! Welcome to PF! :wink:
crazy_v said:
Use the given triangle to prove that for 0 < X <90, 1 + tan^2X = 1 / cos^2X

tanX = b/a
cosX = a/c

But you're there

just put b/a and a/c into the original equation, and you have … ? :smile:
 
Since [tex]\cos \theta=\frac{a}{c} \Rightarrow \cos^2 \theta=\frac{a^2}{c^2}[/tex]. Write c^2 in terms of a and b now, hint Pythagoras.
 
thanks guys

yeah that looks a lot more obvious now, thanks anyways
 

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