Question

$\huge\color{pink}\mathfrak{Answer}$

$\sf{Prove\:That\: \frac{1 + secA}{secA} = \frac{ {sin}^{2}A }{1 - cosA}}$


PLZ ANSWER ASAP ​

Answer 1

$\sf{To\:Prove} \\ \bold {\: \frac{1 + secA}{secA} = \frac{ {sin}^{2}A }{1 - cosA}}$

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$\bold{LHS =\frac{1 + secA}{secA} }$

and

$\bold{RHS =\frac{ {sin}^{2}A }{1 - cosA} }$

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first we solve LHS

$\bold{LHS =\frac{1 + secA}{secA} } \\ \\ \bold{ \implies \: LHS = \frac{1}{sec A} + \cancel\frac{sec A}{sec A} } \\ \\ \bold{\implies \: LHS = \frac{1}{sec A} + 1} \: \: \: \: \: \: \: \: \: \:$

and we know that,

$\boxed{ \bold{ \frac{1}{sec A} = cos A }}$

$\bold{ \implies LHS = \: cos A + 1 } \: \: \: \: \: \: \: \: \: \:$

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now, we solve RHS

$\bold{RHS = \frac{ {sin}^{2}A }{1 - cosA} }$

and we know that,

$\boxed{ \bold{ {sin}^{2}A = 1 - {cos}^{2} A}}$

and

$\bold{1 - {cos}^{2} A = (1 - cosA)(1 + cosA)}$

$\bold{ \implies RHS = \frac{ \cancel{(1 - cosA)}(1 + cosA)} { \cancel{1 - cosA} }} \\ \\ \bold{\implies \: RHS = 1 + cosA} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

so, LHS = RHS

hence, proved

Answer 2

$\color{red}\begin{gathered}\sf{To\:Prove} \\ \bold {\: \frac{1 + secA}{secA} = \frac{ {sin}^{2}A }{1 - cosA}} \end{gathered}$

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$\color{blue}\begin{gathered} \bold{LHS =\frac{1 + secA}{secA} } \\ \end{gathered}$

$\begin{gathered} \bold{RHS =\frac{ {sin}^{2}A }{1 - cosA} } \\ \end{gathered}$

__________________________

first we solve LHS

$\begin{gathered}\bold{LHS =\frac{1 + secA}{secA} } \\ \\ \bold{ \implies \: LHS = \frac{1}{sec A} + \cancel\frac{sec A}{sec A} } \\ \\ \bold{\implies \: LHS = \frac{1}{sec A} + 1} \: \: \: \: \: \: \: \: \: \: \end{gathered}$

and we know that,

$\boxed{ \bold{ \frac{1}{sec A} = cos A }}$

$\color{red}\bold{\implies LHS=\:cos A+1}\:⟹LHS=cosA+1$

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now, we solve RHS

$\begin{gathered} \bold{RHS = \frac{{sin}^{2}A }{1 - cosA} } \\ \end{gathered}$

and we know that,

$\color{red}\boxed{\bold{ {sin}^{2}A = 1 - {cos}^{2} A}}$

$\color{aqua}\bold{1 -{cos}^{2} A = (1 - cosA)(1 + cosA)}\begin{gathered} \bold{ \implies RHS = \frac{ \cancel{(1 - cosA)}(1 + cosA)} { \cancel{1 -cosA} }} \\ \\ \bold{\implies \: RHS = 1 + cosA} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \end{gathered}$

(1−cosA)

⟹RHS=1+cosA

so, LHS = RHS

hence, proved