Question

1/1+costheta+1/1-costheta=cosec^2theta
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Answer 1

Answer:

You are looking to prove the following: 1/(1-cosx) + 1/(1+cosx) = 2csc2x

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 - cos2x = 1 - cos2x

Replace the denominators in both terms by this simplified term:

[(1+cosx) / (1-cos2x)] + [(1-cosx) / (1-cos2x)]

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

[(1+cosx)+(1-cosx)] / [(1-cos2x)]

= (1+cosx+1-cosx) / (1-cos2x)

= 2/(1-cos2x)

Recall the pythagorean identity: sin2x + cos2x = 1

Subtracting cos2x from both sides of the equation, we arrive at: sin2x = 1 - cos2x

Replace 1 - cos2x by sin2x :

2/(1-cos2x) = 2/sin2x

= 2*(1/sin2x)

Recall the following trig identity: 1/sinx = cscx ; likewise, 1/sin2x = csc2x

Replace 1/sin2x with csc2x :

2*(1/sin2x) = 2*(csc2x) = 2csc2x

Thus, we have proven that

1/(1-cosx) + 1/(1+cosx) = 2csc2x