Review the proof of [tex]$\tan (A-B)=\frac{\tan A-\tan B}{1+(\tan A)(\tan B)}$[/tex].
[tex]$\tan (A-B)=[/tex]
Step 1: [tex]$=\frac{\sin (A-B)}{\cos (A-B)}$[/tex]
Step 2: [tex]$=\frac{\sin A \cos B-\cos A \sin B}{\cos A \cos B+\sin A \sin B}$[/tex]
Step 4: [tex]$=\frac{\tan A-\tan B}{1+(\tan A)(\tan B)}$[/tex]
To complete step 3, which expression must fill in each blank space?
[tex]$\cos (A) \cos (B)$[/tex]
[tex]$\cos (A) \sin (B)$[/tex]
[tex]$\sin (A) \cos (B)$[/tex]
[tex]$\sin (A) \sin (B)$[/tex]
Answer (1)
To prove the identity tan ( A − B ) = 1 + ( t a n A ) ( t a n B ) t a n A − t a n B , let's break down the steps in detail to understand how to derive this result. We are given that:
tan ( A − B ) = c o s ( A − B ) s i n ( A − B ) .
Substitute the angle subtraction formulas for sine and cosine in step 1: = c o s A c o s B + s i n A s i n B s i n A c o s B − c o s A s i n B .
We need to simplify this expression to match the form 1 + ( t a n A ) ( t a n B ) t a n A − t a n B . The key step involves dividing both the numerator and denominator by cos A cos B , which is the common term that helps us simplify the expression using tangent.
Divide each term in the numerator and the denominator of step 2 by cos A cos B :
Numerator: c o s A c o s B s i n A c o s B − c o s A c o s B c o s A s i n B = c o s A s i n A − c o s B s i n B = tan A − tan B .
Denominator: c o s A c o s B c o s A c o s B + c o s A c o s B s i n A s i n B = 1 + tan A tan B .
Thus, step 3 gives us:
1 + ( t a n A ) ( t a n B ) t a n A − t a n B .
So, the expression that fills in the blank for step 3 is achieved by dividing both numerator and denominator by cos A cos B , giving us the reduced form involving tangent.
