Question

State True or False.

$\tan \theta -\tan 2\theta -\tan 3\theta$ = (tan$\theta$)($\tan 2\theta$)($\tan 3\theta$)

• A. True
• B. False

Answer 1

The correct option is A True

$\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }$

$\tan 3\theta =\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }$

LHS

$\tan \theta -\tan 2\theta -3\theta$

$\tan \theta -\frac{2\tan \theta }{1-{\tan }^2\theta }-\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }$

$\frac{\tan \theta (1-{\tan }^2\theta )(1-3{\tan }^2\theta )-2\tan \theta (1-3{\tan }^2\theta )-(3\tan \theta -{\tan }^3\theta )(1-{\tan }^2\theta )}{(1-{\tan }^2\theta )(1-3{\tan }^2\theta )}$

$\frac{\tan \theta -{\tan }^3\theta -3{\tan }^3\theta +3{\tan }^5\theta -2\tan \theta +6{\tan }^3\theta -3\tan \theta +{\tan }^3\theta +3{\tan }^3\theta -{\tan }^5\theta }{1-3{\tan }^2\theta -{\tan }^2\theta +3{\tan }^4\theta }$

$\frac{2{\tan }^5\theta +6{\tan }^3\theta -4\tan \theta }{1-4{\tan }^2\theta +3{\tan }^4\theta }\cdots \left(1\right)$

RHS

$\tan \theta \tan 2\theta \tan 3\theta$

$\tan \theta \times \left(\frac{2\tan \theta }{1-{\tan }^2\theta }\right)\times \left(\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }\right)$

$⟹\frac{2{\tan }^2\theta (3\tan \theta -{\tan }^3\theta )}{(1-{\tan }^2\theta )(1-3{\tan }^2\theta )}$

$⟹\frac{6{\tan }^3\theta -2{\tan }^5\theta }{1-4{\tan }^2\theta +3{\tan }^4\theta }\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)⟹$ False