Question

1-sinA/cos A=cosA/1+sinA​

Answer 1

Answer:

LHS = sinA-cosA+1/sinA+cosA-1

divide both numerator and denominator by cosA

LHS=(tanA−1+secA)/(tanA+1−secA)

Now

sec2A=1+tan2A

sec2A−tan2A=1

Using above relation at denominator of LHS

LHS=(tanA−1+secA)/(tanA−secA+sec2A−tan2A)

LHS=(tanA−1+secA)/((secA−tanA)(−1+secA+tanA))

LHS=1/(secA−tanA)

LHS=RHS

Hence Proved.

I think above proof will clear your doubt,

All the best.

KKG.

Step-by-step explanation:

Answer 2

$\sf{\underline{\boxed{\purple{\large{\bold{ Solution }}}}}}$

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$\sf :\implies\:{\dfrac{1 - SinA}{CosA} = \dfrac{ CosA}{ 1 + SinA } }$

⠀⠀⠀⠀

$\sf :\implies\:{\dfrac{1 - SinA}{CosA} \times\dfrac{ 1 + SinA}{ 1 + SinA} = \dfrac{ CosA}{ 1 + SinA } }$

⠀⠀⠀⠀

$\sf{\red{\boxed{\bold{(a + b ) ( a - b ) = a^2 - b^2}}}}$

⠀⠀⠀⠀

$\sf :\implies\:{\dfrac{1^2 - Sin^2A}{(CosA)( 1 + SinA)} = \dfrac{ CosA}{ 1 + SinA } }$

⠀⠀⠀⠀

$\sf{\red{\boxed{\bold{1 - Sin^2A = Cos^2A}}}}$

⠀⠀⠀⠀

$\sf :\implies\:{\dfrac{Cos^2A}{(CosA)( 1 + SinA)} = \dfrac{ CosA}{ 1 + SinA } }$

⠀⠀⠀⠀

$\sf :\implies\:{\dfrac{Cos^2A}{CosA + SinACosA} = \dfrac{ CosA}{ 1 + SinA } }$

⠀⠀⠀⠀

$\sf :\implies\:{\dfrac{Cos^2A}{CosA ( 1 + SinA) } = \dfrac{ CosA}{ 1 + SinA } }$

⠀⠀⠀⠀

$\sf :\implies\:{\dfrac{CosA\times CosA}{CosA ( 1 + SinA) } = \dfrac{ CosA}{ 1 + SinA } }$

⠀⠀⠀⠀

$\sf :\implies\:{\dfrac{CosA}{ 1 + SinA } = \dfrac{ CosA}{ 1 + SinA } }$

⠀⠀⠀⠀

LHS = RHS

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀....Hence Proved