Question

prove thatv1+seca/cosa =sin^2a/1-cosa​

Answer 1

$\\ \\ \large\underline{ \underline{ \sf{ \red{correct \: question:} }}}$

$\large \sf \frac{1 + sec \alpha }{sec \alpha } = \frac{ {sin}^{2} \alpha }{1 - cos \alpha }$

$\\ \\ \large\underline{ \underline{ \sf{ \red{solution:} }}}$

$\\ \sf \: r.h.s = \frac{ {sin}^{2} \alpha }{1 - cos \alpha } \\ \\ \\ \bigstar \boxed{ \bf \: {sin}^{2} \alpha + {cos}^{2} \alpha = 1 } \\ \\ \\ \sf \: \implies \frac{1 - {cos}^{2} \alpha }{1 - cos \alpha } \\ \\ \\ \implies \sf \: \frac{( \cancel{1 - cos \alpha} )(1 + cos \alpha )}{ \cancel{1 - cos \alpha } } \\ \\ \\ \implies \sf \: \pink{1 + cos \alpha \: \: \: \: \: \: \: \: \: \: (1)}$

$\\ \\ \\ \sf \: l.h.s = \frac{1 + sec \alpha }{sec \alpha } \\ \\ \\ \bigstar \boxed{ \bf \:sec \alpha = \frac{1}{cos \alpha } } \\ \\ \\ \implies \sf \: \frac{1 + \frac{1}{cos \alpha } }{ \frac{1}{cos \alpha } } \\ \\ \\ \implies \sf \: \frac{ \frac{cos \alpha + 1}{ \cancel{cos \alpha} } }{ \frac{1}{ \cancel{cos \alpha }} } \\ \\ \\ \implies \sf \: \pink{cos \alpha + 1} \: \: \: \: \: \: \: \: \: \: (2)$

Hence , LHS = RHS (proved)