Trigonometric Functions — MATHS STD 11 Science — Question

1. $\frac{\sin2\beta}{5-\cos2\beta}$

Solution:

Given:

$2\tan\alpha=3\tan\beta$

Now,

$\tan(\alpha+\beta)=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$

$=\frac{\frac{3}{2}\tan\beta-\tan\beta}{1+\Big(\frac{3}{2}\tan\beta\Big)\tan\beta}$

$=\frac{3\tan\beta-2\tan\beta}{2+3\tan^2\beta}$

$=\frac{\tan\beta}{2+3\tan^2\beta}$

$=\frac{\frac{\sin\beta}{\cos\beta}}{2+3\frac{\sin^2\beta}{\cos^2\beta}}$

$=\frac{\sin\beta\cos\beta}{2\cos\beta+3\sin\beta}$

$=\frac{\sin\beta\cos\beta}{2+\sin^2\beta}$

$=\frac{2\sin\beta\cos\beta}{4+2-\cos2\beta}$

$=\frac{\sin2\beta}{4+1-cos2\beta}$

$\therefore\tan(\alpha+\beta)=\frac{\sin2\beta}{5-\cos2\beta}$