Question

Prove that ( cos A – sin A +1) / ( cos A + sin A -1) = cosec A + cot A using the identity
1 + cot2 A = cosec2 A

Answer 1

[tex]\mathfrak{\huge{Answer:-}} \\ \frac{ cosA- sinA+ 1}{cosA + sin A- 1} = cosec A+ cotA \\ \frac{cos A+ (1 - sinA)}{cos A- (1 - sinA)} \times \frac{ cosA+ (1 -sin A)}{ cosA+ (1 - sinA)} = cosec A+ cotA \\ \frac{ {[ cosA+ (1 - sinA)]}^{2} }{ {(cosA)}^{2} - {(1 - sinA)}^{2}} = cosec A+ cotA \\ \frac{ {cos}^{2} A + {(1 - sinA)}^{2} + 2cosA(1 -sinA )}{ {cos}^{2}A - {(1 -sinA )}^{2} } = cosec A+ cotA \\ \frac{ {cos}^{2} A + 1 + {sin}^{2}A - 2sinA + 2cosA - 2cosAsinA }{ {cos}^{2} A - 1 - {sin}^{2} A + 2sinA} = cosec A+ cotA \\ \frac{ {cos}^{2}A + {sin}^{2} A + 1 - 2sinA + 2cos A- 2cosAsin A}{ - {sin}^{2}A - {sin}^{2}A + 2sin A} =cosec A+ cotA \\ \frac{1 + 1 - 2sin A+ 2cosA - 2sinAcosA}{ - 2 {sin}^{2}A + 2 sinA} =cosec A+ cotA \\ \frac{2 - 2sinA + 2cosA - 2sinAcosA}{ 2 {sin}^{2}A - 2sin A} = cosec A+ cotA \\ \frac{2(1 - sinA+ cosA - sinAcosA)}{2({sin}^{2}A - sinA ) } = cosec A+ cotA \\ \frac{1 - sinA+ cosA-sinAcosA }{ {sin}^{2}A - sinA} =cosec A+ cotA \\ \frac{(1 -sinA ) + cosA(1 - sinA)}{sinA(1 - sinA)} = cosec A+ cotA \\ \frac{(1 -sinA)(1 +cosA )}{sinA(1 - sinA )} = cosec A+ cotA \\ \frac{1 +cosA}{sinA} = cosec A+ cotA \\ \frac{1}{sinA} + \frac{cosA}{sinA} =cosec A+ cotA \\ cosec A+ cotA = cosec A+ cotA\\ LHS=RHS\\Hence\: proved [/tex]