If $\frac{\sqrt{2} \sin \alpha}{\sqrt{1+\cos 2 \alpha}}=\frac{1}{7}$ and $\sqrt{\frac{1-\cos 2 \beta}{2}}=\frac{1}{\sqrt{10}}, \alpha, \beta \in\left(0, \frac{\pi}{2}\right)$, then $\tan (\alpha+2 \beta)$ is equal to ______.

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Trigonometric Ratios and Identities

\frac{2 s i n α}{1 + c o s 2 α} = \frac{1}{7}

and

\frac{1 - c o s 2 β}{2} = \frac{1}{10}, α, β \in (0, \frac{π}{2})

, then

tan (α + 2 β)

is equal to ______.

Answer: 1

Since

1 + cos 2 α = 2 cos^{2} α

, the first expression is

\frac{2 sin α}{2 ∣ cos α ∣} = tan α = \frac{1}{7}

for

α

in the first quadrant. Since

\frac{1 - cos 2 β}{2} = sin^{2} β

, the second gives

sin β = \frac{1}{10}

, so

cos β = \frac{3}{10}

and

tan β = \frac{1}{3}

. Then

tan 2 β = \frac{2 \cdot \frac{1}{3}}{1 - \frac{1}{9}} = \frac{3}{4}

, and

tan (α + 2 β) = \frac{\frac{1}{7} + \frac{3}{4}}{1 - \frac{1}{7} \cdot \frac{3}{4}} = \frac{25/28}{25/28} = 1

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Integer answer