Question

$cot \: a + cosec \: a - 1 \div cot \: a - cosec \: a + 1 = 1 + cos \: a \div sin \: a$
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Answer 1

ANSWER:

${ \rm{\frac{cot \: a + cosec \: a - 1}{cot \: a - cosec \: a + 1} = \frac{1 + cos \: a}{sin \: a} }}$

Taking LHS

We know that

cosec²A - cot²A = 1

Let us substitute 1 = cosec²A - cot²A

${ \rm{ \frac{cot \: a + cosec \: a - ( {cosec}^{2}a - {cot}^{2}a) }{cot \: a - cosec \: a \: + 1} }}$

Using

a² - b² = (a + b)(a - b)

$= { \rm{ \frac{cot \: a + cosec \: a - [(cosec \: a + cot \: a)(cosec \: a-cot \: a)] }{cot \: a \: - cosec \: a + 1} }}$

Taking common cotA + cosecA

${\rm{ = \frac{(cosec \: a + cot \: a) \cancel{(1 - cosec \: a + \: cot \: a)}}{ \cancel{cot \: a - cosec \: a + 1}} }}$

Using

cotA = cosA/sinA

cosecA = 1/sinA

${ \rm{ = \frac{cos \: a}{sin \: a} + \frac{1}{sin \: a} = \frac{1 + cos \: a}{sin \: a} }}$

LHS = RHS

Hence proved✓✓