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Math Help - 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

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

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

    then find the values of \tan A and \tan B
    Last edited by mr fantastic; July 24th 2009 at 04:42 AM. Reason: Changed title, deleted request written in large font capitals
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  2. #2
    Member SENTINEL4's Avatar
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    tanA=\frac{sinA}{cosA} (1)

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

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

    tanA=\frac{sinA}{cosA}=\frac{psinB}{qcosB} , but it is \frac{sinB}{cosB}=tanB so...
    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.


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


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


    We can write: . \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

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

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

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