Question

$\frac{tan \: a}{sec \: a - 1} + \frac{tan \: a}{sec \: a + 1} = 2cosec \: a$
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Answer 1

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Answer 2

★Given:-

$\displaystyle \leadsto \sf \frac{tanA}{SecA-1} + \frac{tanA}{secA+1} =2cosecA$

★Proof:-

Formula's used:-

$\displaystyle \sf \leadsto tanA = \frac{sinA}{cosA} \\\\\sf \leadsto secA=\frac{1}{cosA} \\\\\leadsto \sf cosecA=\frac{1}{sinA} \\\\\leadsto \sf 1-cos^2A=sin^2A$

Now,

$\displaystyle :\implies \sf \frac{tanA}{SecA-1} + \frac{tanA}{secA+1} =2cosecA\\\\\\:\implies \sf \dfrac{\dfrac{sinA}{cosA} }{\dfrac{1}{cosA-1} } + \dfrac{\dfrac{sinA}{cosA}}{\dfrac{1}{cosA+1} }\\\\\\:\implies \sf \dfrac{\dfrac{sinA}{cosA} }{\dfrac{(1-cosA)}{cosA}} + \dfrac{\dfrac{sinA}{cosA} }{ \dfrac{(1+cosA) }{cosA}}\\\\\\:\implies \sf \frac{sinA}{\cancel{cosA}} \times \frac{\cancel{cosA}}{(1-cosA)} +\frac{sinA}{\cancel{cosA}} \times \frac{\cancel{cosA}}{(1+cosA)}$

$:\implies \sf \dfrac{sinA}{1-cosA} +\dfrac{sinA}{1+cosA}$

$\displaystyle :\implies \sf \frac{sinA(1+cosA) +sinA (1-cosA)}{ (1-cosA) (1+cosA)}\\\\\\:\implies \sf \dfrac{sinA + \cancel{sinAcosA} +sinA -\cancel{sinAcosA}}{1-cos^2A}\\\\\\:\implies \sf \frac{2sinA}{sin^2A}\\\\\\:\implies \sf \frac{2}{sinA} = 2\times \frac{1}{sinA} \\\\\\:\implies \underline{\boxed{\bold{2cosecA}}}$

Hence proved!

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