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Thread: Finding the values of tan from a pair of given equations

  1. #1
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    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 $\displaystyle \tan B$
    Last edited by mr fantastic; Jul 24th 2009 at 04:42 AM. Reason: Changed title, deleted request written in large font capitals
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  2. #2
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    $\displaystyle tanA=\frac{sinA}{cosA}$ (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...
    $\displaystyle tanA=\frac{p}{q}tanB$
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  3. #3
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    Hello, jashansinghal!

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


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

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

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  4. #4
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    Quote Originally Posted by jashansinghal View Post
    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, $\displaystyle \tan A = \tfrac pq\tan B$).
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