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

1 - ( [1-cos^2]/[1+cos] ) = ? cos

1 - (1-cos)(1+cos)/(1+cos) = ? cos

1 - (1-cos) = ? cos

yes
cos = cos :)

To solve the equation 1 - (sin^2x / 1+cosx) = cosx, we need to simplify and manipulate the equation to isolate the variable x.

Let's start by simplifying the left side of the equation:

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

Since sin^2x is the same as (sinx)^2, we can rewrite the equation as:

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

Next, since (1 + cosx) is in the denominator, we can multiply both sides of the equation by the denominator to eliminate the fraction:

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

Simplifying further:

(1 + cosx) - sin^2x = cosx * (1 + cosx)

Now, let's simplify the equation step by step:

1 + cosx - sin^2x = cosx + cos^2x

Rearranging the equation:

1 - sin^2x = cos^2x

Using the trigonometric identity sin^2x + cos^2x = 1:

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

Since both sides of the equation are equal to 1, we can cancel the 1:

-sin^2x = -cos^2x

Dividing both sides of the equation by -1:

sin^2x = cos^2x

Now, take the square root of both sides of the equation:

sinx = ±cosx

To solve for x, we can set up two equations:

1. sinx = cosx

In this case, we can divide both sides of the equation by cosx:

tanx = 1

Taking the inverse tangent of both sides:

x = arctan(1) + kπ, where k is an integer

2. sinx = -cosx

In this case, we can divide both sides of the equation by cosx:

tanx = -1

Taking the inverse tangent of both sides:

x = arctan(-1) + kπ, where k is an integer

Therefore, the solutions for x are given by:

x = arctan(1) + kπ,
x = arctan(-1) + kπ

where k is an integer value that represents the period of the trigonometric functions.

To simplify the left-hand side of the equation, we'll start by working with the denominator.

1 + cos(x) can be written as cos(x) + 1.

Now, let's substitute this back into the equation:

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

Next, let's simplify the left-hand side by using the fact that sin^2(x) = 1 - cos^2(x):

1 - ((1 - cos^2(x)) / (cos(x) + 1)) = cos(x)

Now, let's combine like terms on the left-hand side:

(1 * (cos(x) + 1) - (1 - cos^2(x))) / (cos(x) + 1) = cos(x)

Simplifying further:

(cos(x) + 1 - 1 + cos^2(x)) / (cos(x) + 1) = cos(x)

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

Now, let's simplify the numerator by factoring out cos(x):

cos(x) * (1 + cos(x)) / (cos(x) + 1) = cos(x)

We can cancel out the common factors of (cos(x) + 1) on both the numerator and denominator:

cos(x) = cos(x)

Therefore, the original equation 1 - (sin^2x / (1 + cosx)) = cosx is true for all values of x.