Pythagorean Identity

To show that the given equation is an identity,

start from the left side,
then derive the right side.

First write the left side
(sin θ + cos θ)²/(sin θ).

(sin θ + cos θ)²
= sin² θ + 2 sin θ cos θ + cos² θ

Square of a Sum: (a + b)²

sin² θ + cos² θ = 1

Split the function into two functions.

1/(sin θ) = csc θ

Cosecant: in a Right Triangle

(2 sin θ cos θ)/(sin θ) = 2 cos θ

Let's see what you've solved.
You changed the left side, (sin θ + cos θ)²/(sin θ),
to the right side, csc θ + 2 cos θ.

So the given equation is an identity.

So write
∴ (sin θ + cos θ)²/(sin θ) = csc θ + 2 cos θ.

This is the solution of this example.

cos² θ = 1 - sin² θ

1 - sin² θ
= 1² - sin² θ
= (1 + sin θ)(1 - sin θ)

Factor the Difference of Two Squares: a² - b²

Cancel the common factors (1 - sin θ).

Then
1 + sin θ
is the answer.

First write the left side
sin² θ(1 + tan² θ).

1 + tan² θ = sec² θ

Secant is the reciprocal of cosine.

So sec² θ = 1/(cos² θ).

Combine sin² θ and 1/(cos² θ).

(sin θ)/(cos θ) = tan θ

Quotient Identity

So (sin² θ)/(cos² θ) = tan² θ.

Let's see what you've solved.
You changed the left side, sin² θ(1 + tan² θ),
to the right side, tan² θ.

So the given equation is an identity.

So write
∴ sin² θ(1 + tan² θ) = tan² θ.

This is the solution of this example.