HOW TO PROVE 1+tanA*tanA/2=tanA*cotA/2-1
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HOW TO PROVE 1+tanA*tanA/2=tanA*cotA/2-1
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Taking 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 ....eqn1
Taking 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)
= 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 ....eqn1
Taking 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)
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