Posted by Evelina on Saturday, November 12, 2011 at 10:45am.
How to do this one?
(sina/1cosa) + (1cosa/sina) =
Also, wanted to ask if I did another one right:
1  sin^2a + tg^2a * cos^2a = 1  sin^2a + (sina/cosa)^2 * cos^2a = 1  sin^2a + sin^2a = 1 ???

trig  Steve, Saturday, November 12, 2011 at 7:08pm
That's not what I get.
(1cosa)/sina = tan(a/2)
so, you have
1/tan(a/2) + tan(a/2)
= (1 + tan^2(a/2))/tan(a/2)
= sec^2(a/2)/tan(a/2)
= 1/cos^2(a/2) * cos(a/2)/sin(a/2)
= 1/sin(a/2)cos(a/2) = 2/sina

trig  Adi, Wednesday, July 20, 2016 at 3:33am
Well it goes like this :
sinA/1cosA + 1cosA/sinA
= sin^2A + (1cosA)^2/(1cosA)sinA (cross multiplying)
= sin^2A + cos^2A + 1  2cosA/(1cosA)sinA
= 1 + 1  2cosA/(1cosA)sinA
(sin^2A + cos^2A =1)
= 2  2cosA/(1cosA)sinA
= 2 (1cosA)/(1cosA)sinA
= 2/sinA
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