Question

tanA/1+secA-tana/1-secA=2 coseco​

Answer 1

$taking \: lhs \\ = \frac{tan \: a}{sec \: a + 1} - \frac{tan \: a}{1 - sec \: a} \\ \\ = \frac{tan \: a(1 - se c \: a \:) - tan \: a(1 + sec \: a) }{(1 + sec \: a)(1 -sec \: a) } \\ \\ = \frac{tan \: a(1 - se c\: a - 1 - sec \: a)}{1 - {sec}^{2} \: a } \\ \\ = tan \: a \: \frac{( - 2sec \: a)}{( - {tan}^{2} \: a) } \\ \\ = \frac{tan \: a \times 2sec \: a}{ {tan}^{2} \: a } \\ \\ = \frac{2sec \: a}{tan \: a} \\ \\ = 2 \times \frac{ \frac{1}{cos \: a} }{ \frac{ sin \: a}{cos \: a} } \\ \\ = 2 \times \frac{1}{sin \: a} \\ \\ = 2cosec \: a \\ \\ = rhs$