Question

1+tanA/2/1-tan A/2=secA+tanA​

Answer 1

1+tanATanA/2=secA ,how to prove it?

Answers

LHS

= 1 + [ tan A. tan (A/2) ]

= 1 + [ sin A. sin (A/2) / cos A. cos (A/2) ]

= [ cos A. cos (A/2) + sin A. sin (A/2) ] / [ cos A. cos (A/2) ]

= [ cos ( A - (A/2) ) ] / [ cos A. cos (A/2) ]

= [ cos (A/2) ] / [ cos A. cos (A/2) ]

= 1 / ( cos A )

= sec A ............... (1)

... RHS

= [ tan A. cot (A/2) ] - 1

= [ sin A. cos (A/2) / cos A. sin (A/2) ] - 1

= { [ sin A. cos (A/2) - cos A. sin (A/2) ] / [ cos A. sin (A/2) ] }

= { [ sin ( A - A/2 ) ] / [ cos A. sin (A/2) ] }

= [ sin (A/2) ] / [ cos A. sin (A/2) ]

= 1 / ( cos A )

= sec A

= LHS ...... from (1) ...........

Answer 2

Answer:

secA+tanA

=1/cosA+sinA/cosA

=1+sinA/cosA

=1+2tanA/2/1-tan'2A/2/1-tan'2A/2/1+tan'2A/2

=1+tan'2A/2+2tanA/2/1-tanA/2.1+tanA/2

=1+tanA/2/1+tanA/2