Question

$\huge\color{pink}\mathfrak{Help Needed}$
Prove this
$\frac{ Cos(90 -A) }{sinA} = \frac{sin(90 - A)}{cosA}$
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Answer 1

To prove

$\bold{\frac{ Cos(90 -A) }{sinA} = \frac{sin(90 - A)}{cosA}}$

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$\bold{LHS = {\frac{ Cos(90 -A) }{sinA} }}$

and

$\bold{RHS = \frac{sin(90 - A)}{cosA}}$

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first we solve LHS

$\bold{LHS = {\frac{ Cos(90 -A) }{sinA} }}$

and we know that,

$\boxed{ \bold{Cos(90 -A) = sinA} }$

$\bold{ \implies \: LHS = \cancel\frac{SinA}{SinA} } \\ \\ \bold{ \implies LHS = 1} \: \: \: \: \: \: \: \: \: \: \:$

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now, we solve RHS

$\bold{RHS = \frac{sin(90 - A)}{cosA}}$

and we know that,

$\boxed{ \bold{sin(90 -A) = cosA} }$

$\bold{ \implies RHS = \cancel \frac{cosA}{cosA} } \\ \\ \bold{ \implies \:RHS =1 } \: \: \: \: \: \: \: \:$

so, LHS = RHS

hence, proved