State whether or not the equation is an identity. Prove your statement.

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

To determine whether the equation sin(x)/cos(x) + cos(x)/1 + sin(x) = sec(x) is an identity, we will simplify both sides of the equation and check if they are equal.

Let's begin by simplifying the left-hand side (LHS) of the equation:

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

To add these fractions, we need a common denominator. The common denominator in this case is cos(x). We can rewrite the term cos(x)/1 as (cos(x) * cos(x))/(1 * cos(x)) to have a common denominator.

(sin(x) + cos^2(x))/cos(x) + sin(x)

Next, we can combine the numerators by adding sin(x) and cos^2(x):

(sin(x) + cos^2(x) + cos(x))/cos(x) + sin(x)

Now we have a single fraction on the LHS:

(sin(x) + cos(x) + cos^2(x))/(cos(x)) + sin(x)

To simplify further, let's rewrite cos^2(x) as 1 - sin^2(x), using the trigonometric identity cos^2(x) = 1 - sin^2(x):

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

Next, we can combine like terms by adding sin(x) and sin(x), as well as cos(x) and 1:

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

Now, let's simplify the right-hand side (RHS) of the equation, which is sec(x):

sec(x) = 1/cos(x)

Now, we can compare the LHS and RHS:

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

To simplify the LHS further, let's rewrite sin^2(x) as 1 - cos^2(x), using the trigonometric identity sin^2(x) = 1 - cos^2(x):

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

Simplifying the numerator of the LHS:

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

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

Now, we can simplify further by canceling out cos(x) in the numerator and denominator of the LHS:

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

cos(x) + 1 + 2tan(x) + sin(x) = 1/cos(x)

Now, let's simplify the equation:

1 + sin(x) + cos(x) + 2tan(x) = 1/cos(x)

Since both sides of the equation are equal, we can conclude that the given equation is an identity.

Note: In this proof, we used trigonometric identities and algebraic simplification to transform the equation and check for equality.