Question

If $\frac{m}{n}=\frac{tanA}{tanB}$ , find the value of $\frac{m+n}{m-n}$

• A. tan (A+B)
• B. tan (A-B)
• C.
• D.

Answer 1

The correct option is C

$\frac{m+n}{m-n}=\frac{tanA+tanB}{tanA-tanB}$ This is I definitely not equal to tan (A+B) or tan (A-B). Other two options are in terms of sin (A± B) or cos (A± B)
We know the terms in sin (A± B) and cos (A ± B) are sinA, cosA, cosA and cosB. So we will

convert tanA into $\frac{sinA}{cosA}$ and tanB to $\frac{sinB}{cosB}$
$\Rightarrow$ $\frac{m+n}{m-n}$ = $\frac{sinAcosB+sinBcosA}{sinAcosB-sinBcosA}$

= $\frac{sin(A+B)}{sin(A-B)}$
Key steps/concepts: (1) tanA = $\frac{sinA}{cosA}$
(2) Formula of sin (A±B)