Question

Cos A - Sin A + 1/Cos A + Sin A-1=csc A+cot A​

Answer 1

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Answer 2

$\frac{cos\ A - sin\ A + 1}{cos\ A + sin\ A - 1} = cosec\ A + cot\ A$

Solution:

Given that, we have to prove:

$\frac{cos\ A - sin\ A + 1}{cos\ A + sin\ A - 1} = cosec\ A + cot\ A$

Take the LHS

$\frac{cos\ A - sin\ A + 1}{(cos\ A + sin\ A) - 1} \\\\\\\frac{cos\ A - sin\ A + 1}{(cos\ A + sin\ A) - 1} \times \frac{(cos\ A + sin\ A) + 1}{(cos\ A + sin\ A) + 1}$

We know that,

$(a-b)(a+b) = a^2-b^2$

Therefore,

$\frac{(cos\ A + 1)^2-sin^2\ A}{(cosA + sin\ A)^2 -1}\\\\Expand\ using\ (a+b)^2 = a^2+2ab+b^2\\\\\frac{cos^2A + 2cos\ A + 1-sin^2A}{cos^2A + 2cos\ A sin\ A + sin^2A -1}\\\\We\ know\ that\ 1 - sin^2A = cos^2A\\\\\frac{cos^2A + 2cos\ A +cos^2A}{1 + 2cos\ A sin\ A - 1}\\\\\frac{2cos^2A + 2cos\ A}{2cos\ A sin\ A}\\\\\frac{2cos\ A(cos\ A + 1)}{2cos\ A sin\ A}\\\\\frac{cos\ A + 1}{sin\ A}\\\\\frac{cos\ A }{sin\ A } + \frac{1}{sin\ A}$

$We\ know\ that\ \frac{1}{sin\ A} = cosec\ A\ and\ \frac{cos\ A}{sin\ A} = cot\ A$

$Therefore\\\\cosec\ A + cot\ A$

Thus, LHS = RHS

Thus proved

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