Question

plz solv this ........

Answer 1

(cosA - sinA + 1 ) / (cosA + sinA - 1) = cosecA + cot A

divide LHS by (sinA / sinA)

[sinA(cosA - sinA + 1 )] / sinA [(cosA + sinA - 1)]                                      

= (cosA - sinA + 1 )/sinA  /  (cosA + sinA - 1)/ sinA                                    

= (cotA -1 +cosecA) / ( cotA + 1 - cosecA)

= cotA + cosecA + [cot^2 A - cosec^2 A ] / ( cotA + 1 - cosecA)

= cotA + cosecA + [(cotA+cosecA)(cotA-cosecA) / ( cotA + 1 - cosecA)

= cotA + cosecA { 1 +cotA - cosecA } /( cotA + 1 - cosecA)

= cotA + cosec A

HENCE PROVED

note : in the step 2 :      -1 is written as cot^2 A - cosec^2 A by the formula  1+cot^2 A = cosec^2 A  

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Answer 2

$\text{ \large{ \underline{ \underline{QUESTION}}}}$


$\frac{ \text{cosA - sinA + 1}}{ \text{cosA + sinA - 1}} = \text{cosecA + cotA}$


$\text{ \large{ \underline{ \underline{SOLUTION}}}}$


$\frac{ \text{cosA - sinA + 1}}{ \text{cosA + sinA - 1}} = \text{cosecA + cotA} \\ \\ LHS \\ \\ \implies \: \frac{ \text{cosA - sinA + 1}}{ \text{cosA + sinA - 1}} \\ \\ \boxed{\text{multiplying \: numerator \: and \: denominator \: with \: sinA } }\\ \\ \implies \: \frac{ \text{sinA(cosA - sinA + 1)}}{ \text{sinA(cosA + sinA - 1)}} \\ \\ \implies \: \frac{ \text{sinAcosA - sin ^{2} A + sinA}}{\text{sinA(cosA + sinA - 1)}} \\ \\ \implies \: \frac{ \text{sinAcosA + sinA - (1 - cos ^{2} A}}{ \text{sinA(cosA + sinA - 1)}} \\ \\ \implies \: \frac{ \text{sinA(cosA + 1) - (1 - cosA)(1 + cosA)}}{\text{sinA(cosA + sinA - 1)}} \\ \\ \implies \: \frac{ \text{1 + cosA(sinA + cosA - 1)}}{ \text{sinA(cosA + sinA - 1)}} \\ \\ \implies \: \frac{ \text{1 + cosA \cancel{(sinA + cosA - 1)} }}{\text{sinA \cancel{(sinA + cosA - 1)} }} \\ \\ \implies \: \frac{ 1}{ \text{sinA}} + \frac{ \text{cosA}}{ \text{sinA}} \\ \\ \textsf{as \: we \: know : } \\ \rightarrow \: \ \frac{1}{ \text{sinA} } = \text{cosecA} \\ \\ \rightarrow \: \frac{ \text{cosA}}{ \text{sinA}} = \text{cotA} \\ \\ \textsf{putting \: these \: values \: in : } \\ \\ \frac{ 1}{ \text{sinA}} + \frac{ \text{cosA}}{ \text{sinA}} \\ \\ \implies \: \text{cosecA} \: + \text{cotA} \\ \\ \boxed{ \textsf{hence \: proved}}$