Question

cosA-sinA÷cosA+sinA=sec2A- tan2A​

Answer 1

Answer is in the image

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Answer 2

$\frac{cos A - sin A}{cosA + sinA} = secA \: - tanA$

RHS :-

$sec 2A - tan2A$

$- > \frac{1}{cos2A} - \frac{sin2 A}{cos2A}$

$- > \frac{1 - sin2 A}{cos2A}$

$- > \frac{1 - (2sinA \: cosA)}{( {cos}^{2} A - {sin}^{2} A)}$

$- > \frac{1 - 2sinA \: cosA}{ {cos}^{2} A - {sin}^{2}A}$

$- > \frac{{cos}^{2} A + {sin}^{2}A - 2sinAcosA}{ {cos}^{2}A - {sin}^{2} A }$

$[Putting \: {cosA }^{2} + {sin}^{2} A \: in \: place \: of \: 1 \: because \: {cosA }^{2} + {sin}^{2} A \: = 1 ]$

$- > \frac{ {(cosA - sinA)}^{2} }{(cosA - sinA)(cosA + sinA)}$

$[ Since \: {a}^{2} + {b}^{2} - 2ab = {(a + b)}^{2} and \: {a}^{2} - {b}^{2} = (a - b)(a + b)]$

$- > \frac{(cosA - sinA)(cosA - sinA)}{(cosA - sinA)(cosA + sinA)}$

$[ Since \: {a + b}^{2} = (a + b)(a + b) ]$

$- > \frac{cosA - sinA}{cosA + sinA}$

So , LHS = RHS

Hence Proved !!!

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