Question

plz..... help yaar ♡♡♡♡

Answer 1

CosA-sinA+1/cosA+sinA-1
=(cosA-sinA+1)(cosA+sinA+1)/(cosA+sinA-1)(cosA+sinA+1)
=(cos²A-cosAsinA+cosA+cosAsinA-sin²A+sinA+cosA-sinA+1)/{(cosA+sinA)²-(1)²}
=(cos²A-sin²A+2cosA+1)/(cos²A+2cosAsinA+sin²A-1)
={cos²A+2cosA+(1-sin²A)}/(1+2cosAsinA-1) [∵, sin²A+cos²A=1]
=(cos²A+2cosA+cos²A)/2cosAsinA
=(2cos²A+2cosA)/2cosAsinA
=2cosA(cosA+1)/2cosAsinA
=(cosA+1)/sinA
=cosA/sinA+1/sinA
=cotA+cosecA
=cosecA+cotA (Proved)

Answer 2

$\underline{ \underline{ \mathfrak{answer}}}$

I am taking theta as a...

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

Using the identity....

${cosec}^{2} a = 1 + {cot}^{2} a$

L. H. S

Taking sin a as LCM in both numerator and denomerator

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

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

BECAUSE..

${cosec}^{2} a \: - {cot}^{2} a \: = 1$

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

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

$cosec \: a \: + cot \: a \: = RHS$

$<marquee>$
...✌. MAY IT WILL HELP YOU.✌..