Question

18. (cotA – cosecA)2 = 1 – cosA/ 1 + cosA

Answer 1

$\bold{ \large{ \underline{ \red{To \: prove}}}} \longrightarrow \\ {(cot \alpha - cosec \alpha )}^{2} = \dfrac{1 - cos \alpha }{1 + cos \alpha } \\ \bold{ \large{ \underline{ \red{ Concept \: used}}}} \longrightarrow \\ \boxed{ \pink{{sin}^{2} \theta = 1 - {cos}^{2} \theta}} \\ \boxed{ \pink{{x}^{2} - {y}^{2} = (x + y) \: (x - y)}}$

$\bold{ \large{ \underline{ \red{Proof}}}} \longrightarrow \: LHS \\ = {( \: cot \alpha - cosec \alpha \: )}^{2} \\ = {( \: \dfrac{cos \alpha }{sin \alpha } - \dfrac{1}{sin \alpha } \: ) }^{2} \\ = {( \: \dfrac{cos \alpha - 1}{sin \alpha } \: )}^{2} \\ = {( \: - \dfrac{1 - cos \alpha }{sin \alpha } \: )}^{2} \\ = \dfrac{ {(1 - cos \alpha )}^{2} }{ {sin}^{2} \alpha } \\ = \dfrac{ {(1 - cos \alpha )}^{2} }{1 - {cos}^{2} \alpha } \\ = \dfrac{ {(1 - cos \alpha )}^{2} }{ {(1)}^{2} - {(cos} \alpha ) ^{2} } \\ = \dfrac{ {(1 - cos \alpha )}^{2} }{(1 + cos \alpha ) \: (1 - cos \alpha )} \\ (1 - cos \alpha ) \: is \: cancel \: out \: from \: nuerator \: and \: denominator \\ = \dfrac{1 - cos \alpha }{1 + cos \alpha } \\ = RHS$