Question

Prove the following identity.
sinx/1-cosx + 1+cosx/sinx = 2(1+cosx)/sinx

Answer 1

To prove the given identity, we need to manipulate the left side of the equation until it matches the right side of the equation.

Starting with the left side:

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

To simplify this expression, we need to find a common denominator. The common denominator is sinx(1-cosx), so we can rewrite the expression as:

(sin^2x + (1-cosx)(1+cosx))/[sinx(1-cosx)]

Simplifying the numerator:

(sin^2x + 1-cos^2x)/[sinx(1-cosx)]

Using the Pythagorean identity sin^2x + cos^2x = 1, we can rewrite the numerator as:

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

Simplifying further:

(2 - 2cos^2x)/[sinx(1-cosx)]

Using the identity cos^2x = 1 - sin^2x, we can rewrite the numerator as:

(2 - 2(1 - sin^2x))/[sinx(1-cosx)]

Simplifying again:

(2 - 2 + 2sin^2x)/[sinx(1-cosx)]

(2sin^2x)/[sinx(1-cosx)]

Canceling out sinx from the numerator and denominator:

(2sinx)/[1 - cosx]

Using the identity 1 - cosx = sin^2x/sinx, we can rewrite the denominator as:

(2sinx)/[(sin^2x/sinx)]

Simplifying further:

(2sinx * sinx)/(sin^2x)

2sinx/sinx

2

Therefore, the left side of the equation simplifies to 2, which matches the right side of the equation. Therefore, the given identity is proven.