prove that cos2A/1-sinA=1+sinA?

Answer 1

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

Answer 2

Comes close to the "Bent Air Law" of Wind Energy Aerodynamics, which relates the force developed on a segment of wind turbine blade from the average angle of deflection of the great mass of wind flowing nearby caused by it. (Remember that air is heavy, massive in large quantities - many tons per second near a long blade.) It is F = dCV^2 ( 1 - cos a ) where d is the air density and C is a constant of proportionality. This "law" can be derived from Newton's Law using the trig identity 2 sin^2 (a) = (1 - cos 2a). T-Shirts may soon be available with the Bent Air Law on them and I hope it is not spam in mentioning this.

Answer 3

If it is cos^2(A)

than cos^2(A)/(1-sinA)=1+sinA multiply both sides by (1-sinA)

and you'll get Cos^2(A)= (1-sin A) x (1+sinA)

which is cos ^2 (A) = 1 - Sin^2(A)

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

which is a trig ID

Answer 4

(cos2A)/(1 - sinA) = 1 + sinA
cos(2A) = -sinA^2 + cosA^2 - sinA^2
(cosA^2 - sinA^2)/(1 - sinA) = 1 + sinA
cosA^2 = 1 - sinA^2
((1 - sinA^2) - sinA^2)/(1 - sinA) = 1 + sinA
(1 - sinA^2 - sinA^2)/(1 - sinA) = (1 + sinA)
(1 - 2sinA^2)/(1 - sinA) = 1 + sinA
1 - 2sinA^2 = (1 + sinA)(1 - sinA)
1 - 2sinA^2 = 1 - sinA + sinA - sinA^2
1 - 2sinA^2 = 1 - sinA^2

as you can see they are not exactly the same.

Unless you meant

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

The (1 - sinA)s cancel out

1 + sinA = 1 + sinA

Answer 5

cos(2A) = (cosA)^2 - (sinA)^2 =
(cos A- sinA)(csA + sinA)
When you divide this by (cosA - sinA) you get
(cosA + sinA).
Your statement can't be true.

Maybe you wanted (cosA)^2 / (1 - sinA) =
{1^2 - (sinA)^2} / (1 - sinA) =
(1 + sinA)(1 - sinA) = 1 + sinA
and this is what you wanted, obviously.

Answer 6

(cosA/(a million + sinA)) + ((a million + sinA)/cosA) ((cosA * cosA) + ((a million + sinA)(a million + sinA)))/(cosA * (a million + sinA)) ((cos(A))^2 + (a million + sinA + sinA + (sin(A))^2)/(cosA * (a million + sinA)) ((a million - sin(A)^2) + (a million + 2sinA + sin(A)^2))/(cosA * (a million + sinA)) (a million - sin(A)^2 + a million + 2sinA + sin(A)^2)/(cos(A) * (a million + sinA)) (2 + 2sin(A))/(cos(A) * (a million + sin(A))) 2(a million + sin(A))/(cos(A) * (a million + sin(A))) 2/cosA ANS : 2/cosA or 2sec(A) --------------------------------------... (sec(A) - a million)/(sec(A) + a million) = (a million - cos(A))^2 * csc(A)^2 ((a million/cosA) - a million)/((a million/cosA) + a million) ((a million - cosA)/cosA) / ((a million + cos(A))/cosA) ((a million - cosA)/cosA) * (cosA/(a million + cosA)) (a million - cosA)/(a million + cosA) (a million - cosA)^2 * csc(A)^2 (a million - cosA)(a million - cos(A)) * (a million/sinA)^2 ((a million - cosA)(a million - cosA))/(sin(A)^2) ((a million - cosA)(a million - cosA))/(a million - cos(A)^2) ((a million - cosA)(a million - cosA))/((a million - cosA)(a million + cosA)) one set of the (a million - cosA)s cancel out (a million - cosA)/(a million + cosA) so considering the fact that the two facets equivalent (a million - cosA)/(a million + cosA) (sec(A) - a million)/(sec(A) + a million) = (a million - cos(A))^2 * csc(A)^2

Answer 7

Check the links below!
Good Luck. Out of all dificult maths I've ever done, trigonometry is my least favorite one!

Answer 8

dude, i wud have certainly solved this 4 yrs back but lackin time today....

Answer 9

its true cos I say it's true. OK?

nuff said.

Answer 10

do u think we dont hav any other work to do