# Calculate x if cos x/ (1-sin x) = 1+sin xCalculate x if cos x/ (1-sin x) = 1+sin x

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You need to remember that by fundamental formula of trigonometry, .

You may write such that:

You may write and such that:

Reducing like terms yields:

**Hence, evaluating the general solution of the equation yields .**

Given:

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

multiplying both sides by (1 - sin x):

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

==> cos x = 1 - sin^2x

==> cos x = cos ^x

Dividing both sides by cos x:

cos x = 1

Therefore:

x = 2*n*pi radians

Where n can be any integer including 0.

cosx/(1-sinx) = 1+sinx

To solve for x.

Solution:

cosx /(1-sinx) = 1+sinx. Divide by 1+sinx.

cosx/[1-sinx)(1+sinx) ] = 1

cosx/(1-sin^2x) = 1

cosx/cos^2x = 1, as sin^2+cos^2 =1. Or 1-s-sin^2 = cos^2x.

1/cosx =1.

cosx =1. So

x = 0, Or x = 2npi for n =0 ,1,2 ,....

We are given: cos x/(1- sin x) = 1+sin x

or cos x= (1-sin x)(1+sin x)= 1- (sin x)^2

We also know that (cos x)^2+(sin x)^2=1

Therefore 1- (sin x)^2=(cos x)^2

So cos x=(cos x)^2

This is only possible if cos x=1 or cos x=0

When cos x=0, x= pi/2

When cos x=1, x=0

We'll cross multiply and we'll get:

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

But, from the fundamental formula of trigonometry, we'll have:

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

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

We'll substitute (2) in (1):

cos x = (cos x)^2

We'll subtract cos x both sides:

(cos x)^2 - cos x = 0

We'll factorize:

cos x * (cos x - 1) = 0

We'll put each factor as zero:

cos x = 0

x = +/- arccos 0 + 2*k*pi

**x = +/- (pi/2) + 2*k*pi**

cos x - 1 = 0

cos x = 1

x = +/-arccos 1 + 2*k*pi

**x = 2*k*pi**