Question

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

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) ...........

Answer 2

Proof:

Step-by-step explanation:

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

Hence, proved

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