Ifcos (α+β)=35,sin(α−β)=513and 0<α,β<π4, then tan (2α)=

0<α,β<π/4, cos (α+β)=3/5, sin (α−β)5/13⇒0<α+β<π/2,0<α−β<π/2and tan (α+β)=4/3,tan(α−β)=5/12tan 2α=tan[(α+β)+(α−β)]=tan(α+β)+tan(α−β)1−tan(α+β)tan(α−β)=4/3+5/12'1−(4/3)(5/12)=48+1536−20=6316.

Ifcos (α+β)=35,sin(α−β)=513and 0<α,β<π4, then tan (2α)=

0<α,β<π/4, cos (α+β)=3/5, sin (α−β)5/13⇒0<α+β<π/2,0<α−β<π/2and tan (α+β)=4/3,tan(α−β)=5/12tan 2α=tan[(α+β)+(α−β)]=tan(α+β)+tan(α−β)1−tan(α+β)tan(α−β)=4/3+5/12'1−(4/3)(5/12)=48+1536−20=6316.

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$I f \cos (α + β) = \frac{3}{5}, \sin (α - β) = \frac{5}{13} a n d 0 < α, β < \frac{π}{4}, t h e n \tan (2 α) =$

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63/16

21/16

33/52

63/52

answer is D.

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Detailed Solution

$\begin{array}{l} 0 < α, β < π / 4, \cos (α + β) = 3 / 5, \sin (α - β) 5 / 13 \Rightarrow 0 < α + β < π / 2, 0 < α - β < π / 2 \\ a n d \tan (α + β) = 4 / 3, \tan (α - β) = 5 / 12 \\ \tan 2 α = \tan [(α + β) + (α - β)] = \frac{\tan (α + β) + \tan (α - β)}{1 - t a n (α + β) \tan (α - β)} = \frac{4 / 3 + 5 / 12}{' 1 - (4 / 3) (5 / 12)} = \frac{48 + 15}{36 - 20} = \frac{63}{16} . \end{array}$