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

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The expression $\cot^2(x) - \csc^2(x)$ can be simplified using sine and cosine to show that it equals -1. Initially, it is expressed as $\dfrac{\cos^2(x)}{\sin^2(x)} - \dfrac{1}{\sin^2(x)}$, which simplifies to $\dfrac{\cos^2(x) - 1}{\sin^2(x)}$. Recognizing that $\cos^2(x) - 1$ is equal to $-\sin^2(x)$ leads to the final simplification of $\dfrac{-\sin^2(x)}{\sin^2(x)} = -1$. The discussion highlights the importance of correctly applying the negative sign in the simplification process. Ultimately, the correct result of the expression is confirmed to be -1.
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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