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Question: If \(\cos(\alpha + \beta) = \frac{4}{5},\sin(\alpha - \beta) = \frac{5}{13}\) and \(\alpha,\beta\) l...

If cos⁡(α+β)=45,sin⁡(α−β)=513\cos(\alpha + \beta) = \frac{4}{5},\sin(\alpha - \beta) = \frac{5}{13} and α,β\alpha,\beta lie between 0 and π4,\frac{\pi}{4}, then tan⁡2α=\tan 2\alpha =

A

1663\frac{16}{63}

B

5633\frac{56}{33}

C

2833\frac{28}{33}

D

None of these

Answer

5633\frac{56}{33}

Explanation

Solution

We have cos⁡(α+β)=45\cos(\alpha + \beta) = \frac{4}{5} and sin⁡(α−β)=513\sin(\alpha - \beta) = \frac{5}{13}

⇒sin⁡(α+β)=35\Rightarrow \sin(\alpha + \beta) = \frac{3}{5} and cos⁡(α−β)=1213\cos(\alpha - \beta) = \frac{12}{13}

⇒2α=sin⁡−135+sin⁡−1513\Rightarrow 2\alpha = \sin^{- 1}\frac{3}{5} + \sin^{- 1}\frac{5}{13}

=sin⁡−1[351−25169+5131−925]= \sin^{- 1}\left\lbrack \frac{3}{5}\sqrt{1 - \frac{25}{169}} + \frac{5}{13}\sqrt{1 - \frac{9}{25}} \right\rbrack

⇒2α=sin⁡−1(5665)⇒sin⁡2α=5665\Rightarrow 2\alpha = \sin^{- 1}{}\left( \frac{56}{65} \right) \Rightarrow \sin 2\alpha = \frac{56}{65}

Now, tan⁡2α=sin⁡2αcos⁡2α=56/6533/65=5633\tan 2\alpha = \frac{\sin 2\alpha}{\cos 2\alpha} = \frac{56/65}{33/65} = \frac{56}{33}.