# Finding the values of tan from a pair of given equations

• Jul 24th 2009, 02:06 AM
jashansinghal
Finding the values of tan from a pair of given equations
If

$\displaystyle \frac{\sin A }{\sin B}=p$

$\displaystyle \frac{\cos A}{\cos B}=q$

then find the values of $\displaystyle \tan A$ and $\displaystyle \tan B$
• Jul 24th 2009, 02:46 AM
SENTINEL4
$\displaystyle tanA=\frac{sinA}{cosA}$ (1)

$\displaystyle \frac{sinA}{sinB}=p \Rightarrow sinA=psinB$ and
$\displaystyle \frac{cosA}{cosB}=p \Rightarrow cosA=pcosB$

Replace these two equations at equation (1) and you have

$\displaystyle tanA=\frac{sinA}{cosA}=\frac{psinB}{qcosB}$ , but it is $\displaystyle \frac{sinB}{cosB}=tanB$ so...
$\displaystyle tanA=\frac{p}{q}tanB$
• Jul 24th 2009, 05:57 AM
Soroban
Hello, jashansinghal!

SENTINEL4 is correct.
And there is not enough information for a unique answer.

Quote:

$\displaystyle \text{If }\:\begin{Bmatrix}{\color{blue}(1)} & \dfrac{\sin A }{\sin B}\:=\:p \\ \\[-3mm] {\color{blue}(2)} & \dfrac{\cos A}{\cos B}\:=\:q \end{Bmatrix}\quad \text{then find the values of }\tan A\text{ and }\tan B$

Divide $\displaystyle {\color{blue}(1)}$ by $\displaystyle {\color{blue}[2]}$: . $\displaystyle \frac{\dfrac{\sin A}{\sin B}}{\dfrac{\cos A}{\cos B}} \;=\;\frac{p}{q} \quad\Rightarrow\quad \frac{\dfrac{\sin A}{\cos A}}{\dfrac{\sin B}{\cos B}} \;=\;\frac{p}{q}$ . $\displaystyle \Rightarrow\quad \frac{\tan A}{\tan B} \;=\;\frac{p}{q}$

We have the ratio of $\displaystyle \tan A$ and $\displaystyle \tan B$ . . . and that's the best we can do.

We can write: .$\displaystyle \begin{Bmatrix}\tan A\:=\:kp \\ \tan B \:=\:kq \end{Bmatrix}\;\;\text{for some real number }k \neq 0$

• Jul 24th 2009, 12:41 PM
Opalg
Quote:

Originally Posted by jashansinghal
If

$\displaystyle \frac{\sin A }{\sin B}=p$

$\displaystyle \frac{\cos A}{\cos B}=q$

then find the values of $\displaystyle \tan A$ and $\displaystyle \tan B$

Notice that $\displaystyle \sin A = p\sin B$ and $\displaystyle \cos A = q\cos B$. But $\displaystyle \sin^2A+\cos^2A=1$. Therefore $\displaystyle p^2\sin^2B + q^2\cos^2B = 1$. Divide by $\displaystyle \cos^2B$ to get $\displaystyle p^2\tan^2B + q^2 = \sec^2B = 1 + \tan^2B$, so that $\displaystyle (p^2-1)\tan^2B = 1-q^2$. Hence $\displaystyle \boxed{\tan B = \pm\sqrt{\frac{1-q^2}{p^2-1}}}$. (And as has already been noted, $\displaystyle \tan A = \tfrac pq\tan B$).