Question

Verify the following identity $\tan \left(3x\right)=\frac{3\tan \left(x\right)-{\tan }^3\left(x\right)}{1-3{\tan }^2\left(x\right)}$

Answer 1

Step $1:$ Finding $\tan \left(2x\right)$:

We know that,

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

$\begin{array}{rcl}\therefore \tan \left(2x\right)& =& \tan \left(x+x\right)\\ & =& \frac{\tan \left(x\right)+\tan \left(x\right)}{1-\tan \left(x\right)\tan \left(x\right)}\\ & =& \frac{2\tan \left(x\right)}{1-{\tan }^2\left(x\right)}\end{array}$

Step $2:$ Finding $\tan \left(3x\right)$:

$\begin{array}{rcl}\therefore \tan \left(3x\right)& =& \tan \left(2x+x\right)\\ & =& \frac{\tan \left(2x\right)+\tan \left(x\right)}{1-\tan \left(2x\right)\tan \left(x\right)}\\ & =& \frac{{\displaystyle \frac{2\tan \left(x\right)}{1-{\tan }^2\left(x\right)}}+\tan \left(x\right)}{1-{\displaystyle \frac{2\tan \left(x\right)}{1-{\tan }^2\left(x\right)}}\tan \left(x\right)}\\ & =& \left(\frac{2\tan \left(x\right)}{1-{\tan }^2\left(x\right)}+\tan \left(x\right)\right)÷\left(1-\frac{2\tan \left(x\right)}{1-{\tan }^2\left(x\right)}\tan \left(x\right)\right)\\ & =& \left(\frac{2\tan \left(x\right)+\tan \left(x\right)\times \left(1-{\tan }^2\left(x\right)\right)}{1-{\tan }^2\left(x\right)}\right)÷\left(1-\frac{2{\tan }^2\left(x\right)}{1-{\tan }^2\left(x\right)}\right)\\ & =& \left(\frac{2\tan \left(x\right)+\tan \left(x\right)-{\tan }^3\left(x\right)}{1-{\tan }^2\left(x\right)}\right)÷\left(\frac{1-{\tan }^2\left(x\right)-2{\tan }^2\left(x\right)}{1-{\tan }^2\left(x\right)}\right)\\ & =& \left(\frac{3\tan \left(x\right)-{\tan }^3\left(x\right)}{1-{\tan }^2\left(x\right)}\right)÷\left(\frac{1-3{\tan }^2\left(x\right)}{1-{\tan }^2\left(x\right)}\right)\\ & =& \left(\frac{3\tan \left(x\right)-{\tan }^3\left(x\right)}{\cancel{1-{\tan }^2\left(x\right)}}\right)\times \left(\frac{\cancel{1-{\tan }^2\left(x\right)}}{1-3{\tan }^2\left(x\right)}\right)\\ & =& \frac{3\tan \left(x\right)-{\tan }^3\left(x\right)}{1-3{\tan }^2\left(x\right)}\end{array}$

Hence, the above identity is verified.