Question

1-sin/1+sin=(sec -tan) ²​

Answer 1

Answer:

Hope the above answer helps you

Answer 2

$\huge\underline\mathrm{Correct\:Question-}$

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Prove the identity :

$\tt{\dfrac{1-SinA}{1+SinA}=(SecA-TanA)^2}$

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$\huge\underline\mathrm{Solution-}$

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★ $\tt{\bold{Taking\:RHS-}}$

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: $\implies$ $\tt{(SecA-TanA)^2}$

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★ $\tt{\bold{Putting\:SecA=\dfrac{1}{CosA}\:and\:TanA=\dfrac{SinA}{CosA}}}$

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: $\implies$ $\tt{(\dfrac{1}{CosA}-\dfrac{SinA}{CosA})^2}$

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★ $\tt{\bold{Taking\:LCM-}}$

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: $\implies$ $\tt{(\dfrac{1-SinA}{CosA})^2}$

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: $\implies$ $\tt{\dfrac{(1-SinA)^2}{(CosA)^2}}$

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★ $\tt{\bold{Put\:Cos^2\:A=1-Sin^2\:A}}$

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: $\implies$ $\tt{\dfrac{(1-SinA)(1-SinA)}{1-Sin^2\:A}}$

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★ $\tt{\bold{Using\:(a+b)(a-b)=a^2-b^2}}$

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: $\implies$ $\tt{\dfrac{(\cancel{1-SinA})\:(1-SinA)}{(\cancel{1-SinA})\:(1+SinA)}}$

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: $\implies$ $\tt{\dfrac{1-SinA}{1+SinA}}$

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: $\implies$ LHS

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Hence proved!