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  4. If α and β lie between 0 and (π/4), find tan 2α, given that cos(α+β)=(4/5) and sin(α-β)=(5/13).

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Q. If $\alpha$ and $\beta$ lie between $0$ and $\frac{\pi}{4}$, find $tan\, 2\alpha$, given that $cos\left(\alpha+\beta\right)=\frac{4}{5}$ and $sin\left(\alpha-\beta\right)=\frac{5}{13}$.

Trigonometric Functions

Solution:

Since $\alpha$ and $\beta$ lies between $\theta$ and $\frac{\pi}{4}$ and
$sin\left(\alpha-\beta\right) > 0$, therefore, both $\alpha+\beta$ and $\alpha - \beta$ are positive acute angles.
Now, $sin\left(\alpha-\beta\right)=\frac{5}{13}$
$\Rightarrow cos\left(\alpha -\beta\right)=\sqrt{1-sin^{2}\left(\alpha -\beta\right)}$
$=\sqrt{1-\left(\frac{5}{13}\right)^{2}}=\frac{12}{13}$
and $cos\left(\alpha +\beta\right)=\frac{4}{5}$
$\Rightarrow sin\left(\alpha +\beta\right)=\sqrt{1-cos^{2}\left(\alpha +\beta\right)}$
$=\sqrt{1-\left(\frac{4}{5}\right)^{2}}=\frac{3}{5}$.
Hence, $tan\left(\alpha+\beta\right)=\frac{sin\left(\alpha +\beta\right)}{cos\left(\alpha +\beta\right)}=\frac{3}{4}$
and $tan\left(\alpha -\beta\right)=\frac{sin\left(\alpha -\beta\right)}{cos\left(\alpha -\beta\right)}=\frac{5}{12}$.
Now, $tan\left(2\alpha\right)=tan\left\{\left(\alpha +\beta\right)+\left(\alpha -\beta\right)\right\}$

$=\frac{tan\left(\alpha +\beta\right)+tan\left(\alpha -\beta\right)}{1-tan\left(\alpha +\beta\right)tan\left(\alpha -\beta\right)}$
$=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}\times\frac{5}{12}}$
$=\frac{56}{33}$.