Proving the Trigonometric Identity: SinA/(1-cosA) - (1-cosA)/SinA 2cosecA

In trigonometry, proving identities is a fundamental task that helps establish relationships between trigonometric functions. One such identity is:

[frac{sin A}{1-cos A} - frac{1-cos A}{sin A} 2csc A]

Step-by-Step Proof

To prove the given identity, we will start from the left-hand side (LHS) and simplify it to match the right-hand side (RHS).

Step 1: Combine the fractions

The LHS can be combined into a single fraction:

[frac{sin A}{1 - cos A} - frac{1 - cos A}{sin A} frac{sin^2 A - (1 - cos A)^2}{(1 - cos A)sin A}]

Step 2: Expand the numerator

Let's expand the numerator:

[sin^2 A - (1 - cos A)^2 sin^2 A - (1 - 2cos A cos^2 A) sin^2 A - 1 2cos A - cos^2 A]

Using the Pythagorean identity [sin^2 A cos^2 A 1]:

[sin^2 A cos^2 A 1 Rightarrow sin^2 A - 1 2cos A - cos^2 A 2cos A]

Therefore, the numerator simplifies to:

[2cos A]

Step 3: Substitute back into the fraction

Substituting this back into the fraction:

[frac{2cos A}{(1 - cos A)sin A}]

Step 4: Simplify the expression

We can simplify this by canceling [1 - cos A] (assuming [1 - cos A eq 0]):

[frac{2cos A}{(1 - cos A)sin A} frac{2}{sin A} 2csc A]

Step 5: Rewrite using cosecant

Since [csc A frac{1}{sin A}], the expression can be rewritten as:

[2csc A]

Conclusion

Therefore, we have shown that:

[frac{sin A}{1 - cos A} - frac{1 - cos A}{sin A} 2csc A]

Thus, the identity is proven.

Alternative Solution

Another way to prove the same identity is as follows:

[frac{sin A}{1 - cos A} - frac{1 - cos A}{sin A} frac{sin^2 A - (1 - cos A)^2}{(1 - cos A)sin A} frac{sin^2 A - 1 2cos A - cos^2 A}{(1 - cos A)sin A}]

This simplifies to:

[frac{1 - (1 - 1 2cos A - cos^2 A)}{(1 - cos A)sin A} frac{1 - 1 2cos A}{(1 - cos A)sin A} frac{2cos A}{(1 - cos A)sin A}]

Cancelling out the common terms, we get:

[frac{2cos A}{(1 - cos A)sin A} frac{2}{sin A} 2csc A]

Hence, the identity is shown to be true.

For further exploration, you can also consider:

[text{LHS} frac{sin A}{1 - cos A} - frac{1 - cos A}{sin A}]

After simplifying, it can be shown to equal:

[2csc A text{RHS}]

Therefore, the identity is proved.

See more on trigonometric identities and proof techniques.