Question

trignometry solve it fast​

Answer 1

Given Expression:

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

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SoluTion:

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Taking RHS,

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★ Changing TanA and SecA in Sin and Cos.

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$\longrightarrow$ $\sf{(\dfrac{1}{CosA}\:-\:\dfrac{SinA}{CosA})^2}$

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Taking LCM,

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$\longrightarrow$ $\sf{(\dfrac{1\:-\:SinA}{CosA})^2}$

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Separating the squares,

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$\longrightarrow$ $\sf{\dfrac{(1\:-\:SinA)^2}{Cos^{2} A}}$

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We know that,

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$\large{\boxed{\rm{\green{Cos^{2}\:A\:=\:1\:-\:Sin^2\:A}}}}$

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$\longrightarrow$ $\sf{\dfrac{(1\:-\:SinA)(1\:-\:SinA)}{1\:-\:Sin^2\:A}}$

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We also know that,

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$\large{\boxed{\rm{\red{(\:a^2\:-\:b^2\:)\:=\:(a\:+\:b)(a\:-\:b)}}}}$

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$\longrightarrow$ $\sf{\dfrac{(1\:-\:SinA)(1\:-\:SinA)}{(1\:+\:SinA)(1\:-\:SinA)}}$

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$\longrightarrow$ $\sf{\dfrac{1\:-\:SinA}{1\:+\:SinA}}$

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$\longrightarrow$ LHS

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Hence proved!

Answer 2

$\Large{\underline{\underline{\mathfrak{\bf{Question}}}}}$

Prove that,

$\sf{\:\dfrac{(1-\sin A)}{(1+\sin A)}\:=\:(\sec A - \tan A)^2}$

$\Large{\underline{\underline{\mathfrak{\bf{Solution}}}}}$

First, Take L.H.S,

$\mapsto\sf{\:\dfrac{(1-\sin A)}{(1+\sin A)}}$

$\small\sf{\:\:\:multiply\:by\:(1-\sin A)\:in\:numerator\:and\:denominator} \\ \\ \mapsto\sf{\:\dfrac{(1-\sin A)(1-\sin A)}{(1+\sin A)(1-\sin A)}} \\ \\ \mapsto\sf{\:\:\dfrac{(1 -\sin A)^2}{(1-\sin^2 A)}}$

$\small\sf{\:\:\:\:(1-\sin^2 A)\:=\:\cos^2 A} \\ \\ \mapsto\sf{\:\:\dfrac{(1-\sin A)^2}{\cos^2 A}} \\ \\ \mapsto\sf{\:\:\left(\dfrac{1-\sin^2 A}{\cos A}\right)^2} \\ \\ \mapsto\sf{\:\:\left(\dfrac{1}{\cos^2 A}-\dfrac{\sin^2 A}{\cos^2 A}\right)^2} \\ \\ \small\sf{\:\:\:\dfrac{1}{\cos A}\:=\:\sec A,\:\:\:\dfrac{\cos A}{\sin A}\:=\:\cot A} \\ \\ \mapsto\sf{\red{\:\:(\sec A-\cot A)^2}}$

=R.H.S

That's proved