2006-12-30 21:27:16 · 9 answers · asked by grv_anm 2 in Science & Mathematics ➔ Mathematics
cosa/2/sina/2-sina/2/cosa/2
=cos^2a/2-sin^2a/2/cos/2sina/2
=cosa/(1/2)sina
=2cota
2006-12-30 21:37:17 · answer #1 · answered by raj 7 · 0⤊ 0⤋
cot a/2 - tan a/2
= 1/tan a/2 - 1
= (1- tan^2 a/2)/tan a/2
= 2 tan a
becuause tan (A+B) = (1-tan A tan B)/(Tan A + tan B)
so put A=B = a/2 and get the above
2006-12-31 05:59:35 · answer #2 · answered by Mein Hoon Na 7 · 0⤊ 0⤋
Using the half angle formulas:
cot a/2 - tan a/2
= â[(1 + cos a)/(1 - cos a)] - â[(1 - cos a)/(1 + cos a)]
Multiply the numerator and denominator of each expression by the numerator of that expression
= â[(1 + cos a)²/(1 - cos² a)] - â[(1 - cos a)²/(1 - cos² a)]
= â[(1 + cos a)²/sin² a] - â[(1 - cos a)²/sin² a]
= (1 + cos a)/sin a - (1 - cos a)/sin a
= [(1 + cos a) - (1 - cos a)]/sin a
= 2cos a/sin a = 2cot a
2006-12-31 05:39:34 · answer #3 · answered by Northstar 7 · 0⤊ 0⤋
cot a/2 - tan a/2 = [ cos a/2 /sin a/2] - [sin a/2 /cos a/2]
= [ cos a/2 square - sin a/2 square] / sin a/2* cosa/2
= [ [cosa]/ [ sina] ]*2
=2tana
2006-12-31 06:00:16 · answer #4 · answered by spnchennai 1 · 0⤊ 0⤋
tan=1/cot
cota/2-1/2cota= 2cot a-1/2cot a
2006-12-31 18:22:08 · answer #5 · answered by Anonymous · 0⤊ 0⤋
[cosa/2] / [sina/2] - [sina/2] / [cosa/2]
now simplify it
[[cos^2 a/2]-[sin^2 a/2]] / [sina/2]*[cosa/2]
hope u can solve now
2006-12-31 05:40:40 · answer #6 · answered by question master 2 · 0⤊ 0⤋
[cos(a/2)/sin(a/2)]-[sin(a/2)/cos(a/2)]
= [(cosa/2)^2-(sina/2)^2]/sin(a/2)*cos(a/2)
= 2cos(a)/sin(a)
= 2cot(a).
2006-12-31 05:42:30 · answer #7 · answered by ankit 1 · 0⤊ 0⤋
http://mathworld.wolfram.com/Half-AngleFormulas.html
2006-12-31 05:30:00 · answer #8 · answered by Dashes 6 · 0⤊ 0⤋
2cota
2007-01-01 01:30:39 · answer #9 · answered by Anonymous · 0⤊ 0⤋