Posted by **Evelina** on Saturday, November 12, 2011 at 10:45am.

How to do this one?

(sina/1-cosa) + (1-cosa/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.

(1-cosa)/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/1-cosA + 1-cosA/sinA

= sin^2A + (1-cosA)^2/(1-cosA)sinA (cross multiplying)

= sin^2A + cos^2A + 1 - 2cosA/(1-cosA)sinA

= 1 + 1 - 2cosA/(1-cosA)sinA

(sin^2A + cos^2A =1)

= 2 - 2cosA/(1-cosA)sinA

= 2 (1-cosA)/(1-cosA)sinA

= 2/sinA

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