Let $\cos (\alpha+\beta)=\frac{4}{5}$ and let $\sin (\alpha-\beta)=\frac{5}{13}$, where $0 \leq \alpha, \beta \leq \frac{\pi}{4}$, then $\tan 2 \alpha=$

JEE Main · 2010 · 11 Apr 2010 · Offline · Trigonometric Ratios and Identities

Let

cos (α + β) = \frac{4}{5}

and let

sin (α - β) = \frac{5}{13}

, where

0 \leq α, β \leq \frac{π}{4}

, then

tan 2 α =

Answer: (A)

Because

0 \leq α, β \leq \frac{π}{4}

, the combined angles lie where the tangents are taken directly. From

cos (α + β) = \frac{4}{5}

we get

sin (α + β) = \frac{3}{5}

and

tan (α + β) = \frac{3}{4}

; from

sin (α - β) = \frac{5}{13}

we get

cos (α - β) = \frac{12}{13}

and

tan (α - β) = \frac{5}{12}

. Writing

2 α = (α + β) + (α - β)

and using the tangent addition formula,

tan 2 α = \frac{\frac{3}{4} + \frac{5}{12}}{1 - \frac{3}{4} \cdot \frac{5}{12}} = \frac{\frac{14}{12}}{\frac{33}{48}} = \frac{56}{33}

JEE Main · 2010 · 11 Apr 2010 · Offline · Single correct option