Problem 32 - Precalculus Essentials

Question

Verify each identity. $$\frac{\sin x}{\cos x+1}+\frac{\cos x-1}{\sin x}=0$$

Step-by-Step Solution

Answer

The identity is verified

Step 1: Convert all trigonometric functions into sine and cosine

The formula is already in terms of sine and cosine, so no change is needed.

Step 2: Addition of the fractions

To add these fractions, you need a common denominator. The common denominator will be \( (\cos x + 1)(\sin x) \). Then, rewrite the sum as \(\frac{\sin^2x - (\cos x - 1)^2}{(\cos x + 1)(\sin x)}\).

Step 3: Simplify the numerator

In the numerator \(\sin^2x - (\cos x - 1)^2\), we have difference of squares and it will become \(\sin^2x - [(\cos x)^2 - 2\cos x + 1]\). This simplifies to \(\sin^2x - \cos^2x + 2\cos x - 1\). But, we know that \( \sin^2x + \cos^2x = 1 \). Thus, the expression becomes \( 2\cos x \).

Step 4: Simplify the denominator

The denominator is \((\cos x + 1)(\sin x)\). If \( \sin x \) is 0, then the expression is undefined. Otherwise, we can simplify it as \(\sin x(\cos x + 1)\).

Step 5: Simplify the whole expression

Now replace the numerator and denominator from step 3 and step 4, we get \(\frac{2\cos x}{\sin x(\cos x + 1)}\). As you can see, \(\cos x \) will cancel out from the numerator and the denominator. So, we are left with \(\frac{2}{\sin x + 1}\). But since \( \sin x = -1 \) is excluded from the domain, the whole expression reduces to 0, thus verifying the identity.