Question

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Answer 1

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Using identities and algebra answer is yours.

Answer 2

Question :

To Prove

$\tt \dagger \: \: \: \: \: \dfrac{cos \: A - sin \: A +1 }{cos \: A + sin \: A - 1} = cosec \: A + cot \: A$

Solution :

We have,

$\tt \dagger \: \: \: \: \: \dfrac{cos \: A - sin \: A +1 }{cos \: A + sin \: A - 1} = cosec \: A + cot \: A$

$\rule{90}3$

Taking LHS.

$\tt : \implies \dfrac{cos \: A - sin \: A +1 }{cos \: A + sin \: A - 1}$

Divide Sin A by Denominator, We get.

$\tt : \implies \dfrac{ \dfrac{ cos \: A }{sin \: A}- \dfrac{ sin \: A}{sin \: A} + \dfrac{ 1}{sin \: A}}{\dfrac{ cos \: A }{sin \: A} + \dfrac{ sin \: A}{sin \: A} - \dfrac{ 1}{sin \: A}}$

$\tt : \implies \dfrac{ cot \: A - 1 + cosec \: A }{cot \: A + 1 - cosec \: A}$

$\tt : \implies \dfrac{ cot \: A + cosec \: A - 1}{cot \: A - cosec \: A + 1}$

$\tt : \implies \dfrac{ cot \: A + cosec \: A - ({cosec }^{2} \: A - {cot}^{2} \: A)}{cot \: A - cosec \: A + 1}$

$\tt : \implies \dfrac{ cot \: A + cosec \: A - (cosec \: A - cot \: A)(cosec \: A + cot \: A)}{cot \: A - cosec \: A + 1}$

$\tt : \implies \dfrac{ cot \: A + cosec \: A \Big[1 - (cosec \: A - cot \: A)\Big]}{cot \: A - cosec \: A + 1}$

$\tt : \implies \dfrac{ cot \: A + cosec \: A(1 - cosec \: A + cot \: A)}{cot \: A - cosec \: A + 1}$

$\tt : \implies \dfrac{ cot \: A + cosec \: A( cot \: A - cosec \: A + 1 )}{cot \: A - cosec \: A + 1}$

$\tt : \implies \dfrac{ cot \: A + cosec \: A \cancel{ (cot \: A - cosec \: A + 1 )}}{ \cancel{cot \: A - cosec \: A + 1}}$

$\tt : \implies cot \: A + cosec \: A.$

Hance Proved.