Question

The value of $\tan 5\mathrm{\theta }$ is.

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

Answer 1

The correct option is D

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

Explanation for the correct option:
Find the value of $\mathbf{tan}\mathbf{5}\mathbf{\theta }$:

$\begin{array}{rcl}& \Rightarrow & \tan 5\mathrm{\theta }=\tan \left(2\mathrm{\theta }+3\mathrm{\theta }\right)\\ & =& \frac{\tan 2\mathrm{\theta }+\tan 3\mathrm{\theta }}{1-\tan 2\mathrm{\theta tan}3\mathrm{\theta }}\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{tan}\mathbf{(}\mathbf{A}\mathbf{+}\mathbf{B}\mathbf{)}\mathbf{=}\frac{\mathbf{tanA}\mathbf{+}\mathbf{tanB}}{\mathbf{1}\mathbf{-}\mathbf{tanAtanB}}\mathbf{]}\\ & =& \frac{{\displaystyle \frac{2\mathrm{tan\theta }}{1-{\tan }^2\mathrm{\theta }}}+{\displaystyle \frac{3\mathrm{tan\theta }-{\tan }^3\mathrm{\theta }}{1-3{\tan }^2\mathrm{\theta }}}}{1-\left({\displaystyle \frac{2\mathrm{tan\theta }}{1-{\tan }^2\mathrm{\theta }}}\right)\left({\displaystyle \frac{3\mathrm{tan\theta }-{\tan }^3\mathrm{\theta }}{1-3{\tan }^2\mathrm{\theta }}}\right)}\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{tan}\mathbf{\left(}\mathbf{2}\mathbf{A}\mathbf{\right)}\mathbf{=}\frac{\mathbf{2}\mathbf{tanA}}{\mathbf{1}\mathbf{-}{\mathbf{tan}}^{\mathbf{2}}\mathbf{A}}\mathbf{,}\mathbf{}\mathbf{tan}\mathbf{3}\mathbf{A}\mathbf{=}\frac{\mathbf{3}\mathbf{tanA}\mathbf{-}{\mathbf{tan}}^{\mathbf{3}}\mathbf{A}}{\mathbf{1}\mathbf{-}\mathbf{3}{\mathbf{tan}}^{\mathbf{2}}\mathbf{A}}\mathbf{]}\\ & =& \frac{{\displaystyle \frac{2\mathrm{tan\theta }\left(1-3{\tan }^2\mathrm{\theta }\right)+\left(3\mathrm{tan\theta }-{\tan }^3\mathrm{\theta }\right)\left(1-{\tan }^2\mathrm{\theta }\right)}{\left(1-{\tan }^2\mathrm{\theta }\right)\left(1-3{\tan }^2\mathrm{\theta }\right)}}}{{\displaystyle \frac{\left(1-{\tan }^2\mathrm{\theta }\right)\left(1-3{\tan }^2\mathrm{\theta }\right)-\left(2\mathrm{tan\theta }\right)\left(3\mathrm{tan\theta }-{\tan }^3\mathrm{\theta }\right)}{\left(1-{\tan }^2\mathrm{\theta }\right)\left(1-3{\tan }^2\mathrm{\theta }\right)}}}\\ & =& \frac{2\mathrm{tan\theta }-6{\tan }^3\mathrm{\theta }+3\mathrm{tan\theta }-3{\tan }^3\mathrm{\theta }-{\tan }^3\mathrm{\theta }+{\tan }^5\mathrm{\theta }}{1-{\tan }^2\mathrm{\theta }-3{\tan }^2\mathrm{\theta }+3{\tan }^4\mathrm{\theta }-6{\tan }^2\mathrm{\theta }+2{\tan }^4\mathrm{\theta }}\\ & =& \frac{{\tan }^5\mathrm{\theta }-10{\tan }^3\mathrm{\theta }+5\mathrm{tan\theta }}{1-10{\tan }^2\mathrm{\theta }+5{\tan }^4\mathrm{\theta }}\end{array}$

Hence, the correct option is D.