Let cos⁡(α+β)=45 and let sin⁡(α−β)=513, where 0

A
19/12
B
20/7
C
25/16
D
56/33

Solution:

We have,

$\begin{array}{l} \cos (α + β) = \frac{4}{5} and \sin (α - β) = \frac{5}{13} \\ \Rightarrow \tan (α + β) = \frac{3}{5} and \tan (α - β) = \frac{5}{12} \\ ∴ \tan 2 α = \tan (α + β + α - β) = \frac{\tan (α + β) + \tan (α - β)}{1 - \tan (α + β) \tan (α - β)} \\ \Rightarrow \tan 2 α = \frac{\frac{3}{5} + \frac{5}{12}}{1 - \frac{3}{5} \times \frac{5}{12}} = \frac{56}{33} \end{array}$

Let $\cos (α + β) = \frac{4}{5}$ and let $\sin (α - β) = \frac{5}{13}$ , where $0 < α, β \leq \frac{π}{4} .$ Then $\tan 2 α =$

Solution:

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