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Chapter 5: Problem 41

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

Short Answer

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

Step by step solution

01

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 \(\frac{\sin\alpha\cos\beta - \cos\alpha\sin\beta}{\cos \alpha \cos \beta}\).
02

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 \(\tan\alpha - \tan\beta\).
03

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.

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Most popular questions from this chapter

Determine whether each statement makes sense or does not make sense, and explain your reasoning.I've noticed that for sine, cosine, and tangent, the trig function for the sum of two angles is not equal to that trig function of the first angle plus that trig function of the second angle.

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