Problem 41 - Precalculus Essentials

Question

Verify each identity. $$\frac{\sin (\alpha-\beta)}{\cos \alpha \cos \beta}=\tan \alpha-\tan \beta$$

Step-by-Step Solution

Answer

The trigonometric identity \(\frac{\sin (\alpha-\beta)}{\cos \alpha \cos \beta} = \tan\alpha - \tan\beta\) is proven to be correct.

Step 1: Rewrite the Left Side

Let's first deal with the left side of the equation. The left side of the given equation is \(\frac{\sin (\alpha-\beta)}{\cos \alpha \cos \beta}\). We know that the difference of two angles in sine can be rewritten using the trigonometric identity \(\sin(\alpha-\beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta\). So, after the substitution the left hand side (l.h.s) becomes \(\frac{\sin\alpha\cos\beta - \cos\alpha\sin\beta}{\cos \alpha \cos \beta}\).

Step 2: Simplify the Left Side

Further simplify the l.h.s by dividing both terms in the numerator by \(\cos \alpha \cos \beta\). So, l.h.s becomes \(\frac{\sin\alpha\cos\beta}{\cos \alpha \cos \beta} -\frac{\cos\alpha\sin\beta}{\cos \alpha \cos \beta}\). After reducing, we have \(\tan\alpha - \tan\beta\).

Step 3: Validate the Identity

So, we have simplified the left hand side to \(\tan\alpha - \tan\beta\), which is exactly the expression on the right hand side of the equation given. Hence, the given identity is verified.