Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

$\frac{m+n}{m-n}=\frac{tanA+tanB}{tanA-tanB}$. This is definitely not equal to tan (A+B) or tan (A-B). Other two options are in terms of $sin(A\mp B)orcos(A\mp B)$

We known the terms in sin $(A\mp B)$ and $cos(A\mp B)$are sin ~, cos A, and cos B. 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) $tanA=\frac{sinA}{cosA}$

$tanB=\frac{sinB}{cosB}$

(2)Formula of sin $(A\mp B)$