$$\csc^2+\sec^2=1?$$
I thought I could just use reciprocal from the other formula $\sin^2+\cos^2=1$, can you explain what's wrong?
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You can't just "use reciprocal", as you put it, because $$\frac{1}{x^2+y^2}$$ is not the same thing as $$\frac{1}{x^2}+\frac{1}{y^2}$$
Zev Chonoles
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OH. i thought they were two separate terms, like how you can "move" cos^2 to the other side so it'd be sin^2=1-cos^2 – John Jul 22 '15 at 12:14
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Your algebra isn't quite right (you need to divide through each individual term in turn).
$$\sin^2x + \cos^2x = 1$$ So if you divide through by $\sin^2x$, you get:
$$1 + \frac{\cos^2x}{\sin^2x} = \frac{1}{\sin^2x}$$
$$ 1 + \text{cot}^2x = \text{cosec}^2x $$
likewise, if you divide through by $\cos^2x$, you obtain:
$$ \frac{\sin^2x}{\cos^2x} +1 = \frac{1}{\cos^2x}$$
$$ \tan^2x + 1 = \sec^2x $$
John_dydx
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We have $sin^{2}x + cos^{2}x = 1$ .Taking reciprocal we get $1 / sin^{2}x + cos^{2}x = 1$ ,then converting sin to cosec and cos to sec and then taking lcm of denominator we get $sec^{2}xcosec^{2}x/sec^{2}x + cosec^{2}x$ = 1
Taylor Ted
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1+1 Taking this one step further we have the 'other' Pythagorean identity: $$\sec^2 x + \csc^2 x = \sec^2 x \csc^2 x$$ – John Joy Jul 22 '15 at 13:03
cscandsec. Nobody knows what they are (I have to recheck every time which is1/sinand which1/cos- big confusion), they are unnecessary, don't naturally appear in any sensible natural laws and, as you have noticed, don't have a nice enough identity connecting them. There's an objective reason that $(\cos x,\sin x)$ are a natural choice: they are components of a unit vector at a certain angle (or components of a complex number $e^{i x}$), and as such, all the nice properties of vectors and complex numbers hold. $\csc$ and $\sec$ are just names for some derived expressions. – orion Jul 22 '15 at 12:17