JEE Main 2020 (Online) 8th January Evening ShiftQuestion 1
If $\frac{\sqrt2\sin\alpha}{\sqrt{1+\cos2\alpha}}=\frac17$ and $\sqrt{\frac{1-\cos2\beta}{2}}=\frac1{\sqrt{10}}$, $\alpha,\beta\in\left(0,\frac\pi2\right)$, then $\tan(\alpha+2\beta)$ is equal to ________.
Trigonometric Ratio & Identities
Answer$1$
Solution
Concept. The two conditions simplify to $\tan\alpha$ and $\sin\beta$. Formulas used. (1) $1+\cos2\alpha=2\cos^2\alpha$, $1-\cos2\beta=2\sin^2\beta$(2) $\tan2\beta=\frac{2\tan\beta}{1-\tan^2\beta}$, $\tan(u+v)=\frac{\tan u+\tan v}{1-\tan u\tan v}$Step 1 - $\tan\alpha$. By (1), $\dfrac{\sqrt2\sin\alpha}{\sqrt2\cos\alpha}=\tan\alpha=\frac17$. Step 2 - $\tan\beta$. By (1), $\sin\beta=\frac{1}{\sqrt{10}}$, so $\tan\beta=\frac13$ and by (2) $\tan2\beta=\frac{2/3}{8/9}=\frac34$. Step 3 - Value. By (2), $$\tan(\alpha+2\beta)=\frac{\frac17+\frac34}{1-\frac{3}{28}}=\frac{25/28}{25/28}=1.$$ Answer: 1