Asked by Evelina
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 ???
(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 ???
Answers
Answered by
Steve
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
(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
Answered by
Adi
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
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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