Question

Prove the following
Cot A-1÷2-sec2A=cotA÷1+tan A

Answer 1

The box in blue marks the equations that i used in solving this question.

Answer 2

$: \rightarrow \underline {\underline \mathfrak{Required \: Answer :-}}$

---------------------

Prove the following :

$\red{ \tt{ { \huge{ ↳}}: \: \: \frac{cot \: A - 1}{2 - {sec}^{2} \: A } = \frac{cot \: A}{1 + tan \: A} }} \\ \\ \\ \tt{L.H.S } : \rightarrow \blue{ \tt{ \: \dfrac{ \frac{cos \:A }{sin \: A} - 1}{1 + 1 - {sec}^{2} A} }} \\ \\ \\ : \rightarrow \blue{ \tt{ \: \frac{ \frac{cos \:A - sin \: A }{sin \:A } }{1 + {tan}^{2}A } }} \\ \\ \\ : \rightarrow \blue{ \tt{ \: \frac{ \frac{cos \:A - sin \: A }{sin \: A} }{1 + \frac{ {sin}^{2}A }{ {cos}^{2}A } } }} \\ \\ \\ : \rightarrow \blue{ \tt{ \: \frac{ \frac{cos \: A - sin \: A}{sin \: A} }{ \frac{ {cos}^{2} A + {sin}^{2} A}{ {cos}^{2} A} } }} \\ \\ \\ : \rightarrow \blue{ \tt{ \: \frac{cos \:A - sin \: A }{sin \: A} \times \frac{ {cos}^{2}A }{ {cos}^{2}A + {sin}^{2} A } }} \\ \\ \\ : \rightarrow \blue{ \tt{ \: \frac{ \cancel{cos \: A - sin \: A}}{sin \:A } \times \frac{ {cos}^{2} A}{(cos \:A + sin \: A) \cancel{(cos \: A - sin \: A)}} }} \\ \\ \\ : \rightarrow \blue{ \tt{ \: \frac{ {cos}^{2}A }{sin \: A(cos \:A + sin \: A)} }} \\ \\ \\ \\ \tt{ R.H.S : \rightarrow } \blue { \tt{ \: \frac{cot \: A}{1 + tan \: A} }} \\ \\ : \rightarrow \blue{ \tt{ \: \frac{ \frac{cos \: A}{sin \: A} }{1 + \frac{sin \: A}{cos \: A} } }} \\ \\ \\ : \rightarrow \blue{ \tt{ \: \frac{ \frac{cos \:A }{sin \: A} }{ \frac{cos \: A + sin \:A }{cos \: A} } }} \\ \\ \\ : \rightarrow \blue{ \tt{ \: \frac{cos \: A}{sin \: A} \times \frac{cos \: A}{ cos \: A + sin \: A} }} \\ \\ \\ : \rightarrow \blue{ \tt{ \: \frac{ {cos}^{2}A }{sin \: A(cos \: A + sin \:A) } }} \\ \\ \therefore \tt{ \: L.H.S = R.H.S \: \: \: \: \: \: \: \: \: \: \: \: (proved)}$

_____________________

@MissSolitary ✌️

_________