Question

Prove that:

1+tan(A)tan(A/2) = tan(A)cot(A/2)-1 = sec(A)

Answer 1

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) ........................................... Q.E.D.
.........................................

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