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Thread: To Find the Value of tan

  1. #1
    Member kjchauhan's Avatar
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    To Find the Value of tan

    Please help me to solve the following :::

    If $\displaystyle \cos (\alpha + \beta) = \frac{4}{5}, \sin(\alpha - \beta) = \frac{5}{13}$ and $\displaystyle \alpha, \beta \in (0, \frac{\pi}{4})$, then find the value of $\displaystyle \tan (2 \alpha)$.

    Thanks in advance..
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  2. #2
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    Quote Originally Posted by kjchauhan View Post
    Please help me to solve the following :::

    If $\displaystyle \cos (\alpha + \beta) = \frac{4}{5}, \sin(\alpha - \beta) = \frac{5}{13}$ and $\displaystyle \alpha, \beta \in (0, \frac{\pi}{4})$, then find the value of $\displaystyle \tan (2 \alpha)$.
    if $\displaystyle \cos(\alpha + \beta) = \frac{4}{5}$ , then $\displaystyle \sin(\alpha + \beta) = \frac{3}{5}$

    if $\displaystyle \sin(\alpha - \beta) = \frac{5}{13}$ , then $\displaystyle \cos(\alpha - \beta) = \frac{12}{13}$

    $\displaystyle \sin(2\alpha) = \sin[(\alpha + \beta) + (\alpha - \beta)] = \sin(\alpha + \beta)\cos(\alpha - \beta) + \cos(\alpha + \beta)\sin(\alpha - \beta)
    $

    $\displaystyle \cos(2\alpha) = \cos[(\alpha + \beta) + (\alpha - \beta)] = \cos(\alpha + \beta)\cos(\alpha - \beta) - \sin(\alpha + \beta)\sin(\alpha - \beta)$

    finally, note that ...

    $\displaystyle \displaystyle \tan(2\alpha) = \frac{\sin(2\alpha)}{\cos(2\alpha)}$
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  3. #3
    Member kjchauhan's Avatar
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    Thank you very much..
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