can anyone give me a solution of (1-sinx)/cosx?thks.?

Answer 1

I = (1-sinx) / cosx

(1-sin x) cos x / cos^2 x

(1-sin x) cos x / 1 - sin^2 x

(1-sin x) cos x / (1 - sin x) (1+sin x)

I = cos x / (1+sin x)

multiply

I^2 = (1-sin x) / (1+sin x)
use 1 = sin^2 x/2+cos^2 x/2 and sin x = 2 sin x/2 cos x/2

I^2 = (sin x/2 - cos x/2)^2 / (sin x/2 + cos x/2)^2

I = (sin x/2 - cos x/2) / (sin x/2 + cos x/2)

Answer 2

Asking for a solution implies an equation, but you have only given us a solution. Did you mean

(1 - sin x)/cos x = 0

If this is the problem, try multiplying both sides by cos x. This starts the following chain of reasoning:

(1 - sin x) = 0

sin x = 1

x = arcsin 1

x = π/2 (and 5π/2, 9π/2, etc. But let's not worry about them.)

However, this solution in not valid, because when we substitute it back into the original equation, we get an indeterminate form:

(1 - sin π/2)/cos π/2 = (1 - 1)/0 = 0/0 ???

If, on the other hand, you wish to just simplify the expression, take the following tack:

(1 - sin x)/cos x = (1/cos x) - (sin x/cos x) = sec x - tan x

Answer 3

A solution would require there was en equal sign somewhere; theres not.

Steve

Answer 4

You should have an equationthen solve it, your statement is not an equation.