Question

[ tan A / ( sec A - 1 ) ] = [ ( sec A + 1 ) / tan A ] . prove it​

Answer 1

Answer:

[ tan A / ( sec A - 1 ) ] = [ ( sec A + 1 ) / tan A ]

LHS=

[ tan A / ( sec A - 1 ) ]

=tanA (SecA+1)/SecA-1(SecA+1). [on rationalising]

=tanA(SecA+1)/tan^2A

=SecA+1/tanA=RHS

Hence Proved