Question

Find the value of $\tan 5\theta$.

• A. $\frac{5\tan \theta -10{\tan }^3\theta +{\tan }^5\theta }{1-10{\tan }^2\theta +5{\tan }^4\theta }$
• B. $\frac{\tan \theta -10{\tan }^3\theta +5{\tan }^5\theta }{1-10{\tan }^2\theta +5{\tan }^4\theta }$
• C. $\frac{5\tan \theta -10{\tan }^3\theta +{\tan }^5\theta }{1-5{\tan }^2\theta +10{\tan }^4\theta }$
• D. $\frac{\tan \theta +10{\tan }^3\theta +5{\tan }^5\theta }{1-10{\tan }^2\theta +5{\tan }^4\theta }$

Answer 1

The correct option is A $\frac{5\tan \theta -10{\tan }^3\theta +{\tan }^5\theta }{1-10{\tan }^2\theta +5{\tan }^4\theta }$

$\tan 5\theta =\tan \left(\tan 3\theta +2\theta \right)=\frac{\tan 3\theta +\tan 2\theta }{1-\tan 3\theta \tan 2\theta }$

$($ Using $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B})$

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

$($ Using $\tan 3x=\frac{3\tan x-{\tan }^{3}x}{1-3{\tan }^{2}x}$ and $\tan 2x=\frac{2\tan x}{1-{\tan }^{2}x})$

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

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

$=\frac{{\tan }^5\theta -10{\tan }^3\theta +5\tan \theta }{5{\tan }^4\theta -10{\tan }^2\theta +1}$