Question

If 2 tan a = 3 tan B then prove that tan(a - b) = sin2B/5- cos 2ß​

Answer 1

Step-by-step explanation:

$\tan( \alpha ) = 3 \tan( \beta ) \div 2$

$\tan( \beta ) = 2 \tan( \alpha ) \div 3$

l

$l.h.s = \tan(a - b)$

$tan(a - b) = \tan(a) - \tan(b) \div (1 + \tan(a) \times \tan(b) )$

$= put \: value. \tan(a) and \tan(b)$

Answer 2

Step-by-step explanatiGiven $2 \tan \alpha=3 \tan \beta$

$\rm\[ \tan \alpha=\dfrac{3}{2} \tan \beta \]$

$\[ \begin{aligned} \rm\text { Now, } \tan (\alpha-\beta)=\frac{\tan \alpha-\tan \beta}{1+\tan \alpha \cdot \tan \beta} =\frac{\dfrac{3}{2} \tan \beta-\tan \beta}{1+\dfrac{3}{2} \tan \beta \tan \beta} \\ \\ \rm =\frac{\tan \beta}{2+3 \tan ^{2} \beta}=\frac{\dfrac{\sin \beta}{\cos \beta}}{2+3 \dfrac{\sin ^{2} \beta}{\cos ^{2} \beta}}=\frac{\sin \beta \cos \beta}{2 \cos ^{2} \beta+3 \sin ^{2} \beta} \\ \\ \rm=\frac{\sin \beta \cos \beta}{1+\cos 2 \beta+\dfrac{3}{2}(1-\cos 2 \beta)}=\frac{2 \sin \beta \cdot \cos \beta}{5+2 \cos 2 \beta-3 \cos 2 \beta}=\frac{\sin 2 \beta}{5-\cos 2 \beta} \end{aligned} \]$