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qn
\[\frac{ 1 + \sin x }{ 1 - \sin x } - \frac{ 1 - \sin x }{ 1 + \sin x } = 4 \tan x \sec x\]
\[\frac{ (1+\sin x)^2 + (1-\sin x)^2 }{ (1+\sin x) (1-\sin x) }\]
sorry that was minus; oops. :P i'll correct it. \[\frac{ 1 + 2\sin x + \sin^2x - (1 - 2\sin x + \sin^2x) }{ 1 - \sin^2x }\]
On simplifying - \[\frac{ 4\sin x }{ \cos^2x } => \frac{ 4 \sin x }{ \cos x } . \frac{ 1 }{ \cos x }\]
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I hope you can do it from here -
thank you!...i can continue from there
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