Prove the Trigonometic identity

You are looking to prove the following: 1/(1-cosx) + 1/(1+cosx) = 2csc^{2}x

To prove that the left hand side of the equation does in fact equal the right hand side of the equation, we start by solving for the left hand side of the equation.

1/(1-cosx) + 1/(1+cosx)

Since we are essentially adding two fractions, we need to make it so that both fractions have a common denominator. To do so, we multiply the first term by: (1+cosx)/(1+cosx), and the second term by: (1-cosx)/(1-cosx). Notice that (1+cosx)/(1+cosx) = 1 and that (1-cosx)/(1-cosx) = 1, so you are not actually altering the value of the terms you are multiplying them by.

[1/(1-cosx) * (1+cosx)/(1+cosx)] + [1/(1+cosx) * (1-cosx)/(1-cosx)]

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

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

Notice that the denominator in the first term, (1-cosx)(1+cosx), is equal to the denominator in the second term, (1+cosx)(1-cosx).

(1-cosx)(1+cosx) = 1 + cosx - cosx - cos^{2}x = **1 - cos**^{2}**x**

Replace the denominators in both terms by this simplified term:

** [(1+cosx) / (1-cos**^{2}**x)] + [(1-cosx) / (1-cos**^{2}**x)]**

Now that both fractions have a common denominator, we can add them:

[(1+cosx)+(1-cosx)] / [(1-cos^{2}x)]

= (1+cosx+1-cosx) / (1-cos^{2}x)

= **2/(1-cos**^{2}**x)**

Recall the pythagorean identity: sin^{2}x + cos^{2}x = 1

Subtracting cos^{2}x from both sides of the equation, we arrive at: sin^{2}x = 1 - cos^{2}x

Replace 1 - cos^{2}x by sin^{2}x :

2/(1-cos^{2}x) = **2/sin**^{2}**x**

= **2*(1/sin**^{2}**x)**

Recall the following trig identity: 1/sinx = cscx ; likewise, 1/sin^{2}x = csc^{2}x

Replace 1/sin^{2}x with csc^{2}x :

2*(1/sin^{2}x) = 2*(csc^{2}x) = **2csc**^{2}**x**

Thus, we have proven that 1/(1-cosx) + 1/(1+cosx) = 2csc^{2}x

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