Simplifying $\cot^2(x)-\csc^2(x)$: 1

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Discussion Overview

The discussion revolves around simplifying the expression $\cot^2(x) - \csc^2(x)$ using sine and cosine. Participants explore different approaches to the simplification and address potential errors in the process.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant simplifies $\cot^2(x) - \csc^2(x)$ to 1 by expressing it in terms of sine and cosine.
  • Another participant points out a missing minus sign in the simplification process.
  • A third participant uses the identity $\sin^2x + \cos^2x = 1$ to derive that $\cot^2(x) - \csc^2(x) = -1$.
  • A later reply agrees with the identification of the missing minus sign and provides a corrected simplification leading to -1.

Areas of Agreement / Disagreement

Participants express differing views on the simplification of the expression, with some arriving at 1 and others concluding it equals -1. The discussion remains unresolved regarding the correct simplification.

Contextual Notes

Participants rely on trigonometric identities and algebraic manipulation, but there are unresolved issues regarding the handling of signs in the simplification process.

karush
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Write
$\cot^2(x)-\csc^2(x)$
In terms of sine and cosine and simplify
So then
$\dfrac{\cos ^2(x)}{\sin^2(x)}
-\dfrac{1}{\sin^2(x)}
=\dfrac{\cos^2(x)-1}{\sin^2(x)}
=\dfrac{\sin^2(x)}{\sin^2(x)}=1$
Really this shrank to 1

Ok did these on cell so...
 
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karush said:
Write
$\cot^2(x)-\csc^2(x)$
In terms of sine and cosine and simplify
So then
$\dfrac{\cos ^2(x)}{\sin^2(x)}
-\dfrac{1}{\sin^2(x)}
=\dfrac{\cos^2(x)-1}{\sin^2(x)}
=\dfrac{\sin^2(x)}{\sin^2(x)}=1$
Really this shrank to 1

Ok did these on cell so...
There is a minus sign missing (can you see where?).
 
$$\sin^2x+\cos^2x=1$$

$$1+\cot^2x=\csc^2x$$

$$\cot^2x-\csc^2x=-1$$
 
Opalg said:
There is a minus sign missing (can you see where?).

ok i think the negative follows thru now
$\dfrac{\cos ^2(x)}{\sin^2(x)}
-\dfrac{1}{\sin^2(x)}
=-\dfrac{\cos^2(x)-1}{\sin^2(x)}
=-\dfrac{\sin^2(x)}{\sin^2(x)}=-1$
 

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