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

Answer 1

Let's focus on consolidating the left hand of our equation first: we have two fractions with the denominators 1-cos(x) and 1+cos(x). The least common denominator of those two is (1-cos(x))(1+cos(x)) = 1 - cos²(x). Before moving forward, let's rename the left-hand-side fractions and combine them:

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

1-cos(x) (1+cos(x)) 1-cos(x) (1-cos(x)) 1-cos²(x) 1-cos²(x)

From the identity sin²(x) + cos² = 1, we can subtract cos²(x) from either side to conclude that sin²(x) = 1-cos²(x). Substituting this into our expression, we now have

2 = 2 · 1

sin²(x) sin²(x)

Since csc(x) = 1/sin(x), csc²(x) = 1/sin²(x), and we can rewrite our expression as 2 csc²(x), completing our proof.

Answer 2

Hey!

In trig proofs we should always pick one side of the equation and make it look like the other. With this problem we should take the LHS and get it to the form 2csc^2x, we should start by trying to combine the two fractions, and we can do this by getting a common denominator...

(\frac{1}{1-cos(x)} * \frac{1+cos(x)}{1+cos(x)}) + (\frac{1}{1+cos(x)} * \frac{1-cos(x)}{1-cos(x)}) =

\frac{1+cos(x)}{1-cos^2(x)} + \frac{1-cos(x)}{1-cos^2(x)}

From here, we can now combine the fractions and should be able to use trig identities to finish the proof (hint: sin^2(x) + cos^2(x) = 1)

Feel free to reach out if you still have questions about the rest of this proof!

EDIT: I didn't realize LaTeX was not supported... here are the two lines of math not using latex notation:

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

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