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Trig identities

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\[\huge \frac{ 1 }{\sin x+1 }- \frac{ 1 }{ sinx-1 } = \frac{ 2 }{ \cos^{2}x }\]
\[\huge \frac{ 1 }{ sinx+1 } - \frac{ 1 }{ sinx-1 } = \frac{ 2 }{ 1-\sin^2x }\]
You're going to have to add the ones on the left before you see it.

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Other answers:

what do you mean ? :S
\[\Large \frac{a}{b} - \frac{c}{d} = \frac{ad-cb}{bd} \]
for the right side i get this : \[\huge \frac{ (sinx-1)-(sinx+1) }{ 1-\sin^2x }\]
dont make your life hard combine the left hand side into 1 fraction first
then multiply the denominators to remove the fraction
then it would be extremely easy to solve
when u make the LHS into one fraction, u will get something like \[\frac{ a }{ b }=\frac{ c }{ d }\] \[ad = cb\]

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