Trigonometric Functions — MATHS STD 11 Science — Question

2. $\sin4\beta$

Solution:

Given that, $\tan\alpha=\frac{1}{7}$ and $\tan\beta=\frac{1}{3}$

$\cos2\alpha=\frac{1-\tan^2\alpha}{1+\tan^2\alpha}=\frac{1-\Big(\frac{1}{7}\Big)^2}{1+\Big(\frac{1}{7}\Big)^2}=\frac{1-\frac{1}{49}}{1+\frac{1}{49}}$

$=\frac{48}{50}=\frac{24}{25}$

Now $\tan2\beta=\frac{2\tan\beta}{1-\tan^2\beta}=\frac{2\times\frac{1}{3}}{1-\frac{1}{9}}=\frac{\frac{2}{3}}{\frac{8}{9}}=\frac{2}{3}\times\frac{9}{8}=\frac{3}{4}$

$\therefore\tan^2\beta=\frac{3}{4}$

$\sin4\beta=\frac{2\tan2\beta}{1+\tan^22\beta}$$$

$=\frac{2\times\frac{3}{4}}{1+\Big(\frac{3}{4}\Big)^2}=\frac{\frac{3}{2}}{1+\frac{9}{16}}=\frac{3}{2}\times\frac{16}{25}=\frac{24}{25}$

$\cos2\alpha=\sin4\beta=\frac{24}{25}$

Hence, the correct option is (b).