+26367 If
\(\cos(x) = \dfrac{2ab}{a^2 + b^2}\),
then find \(\tan\left(\frac{x}{2}\right)^2\) in terms of a and b.
I assume: \(\tan\left(\frac{x}{2}\right)^2 = \tan^2\left(\frac{x}{2}\right)\)
\(\begin{array}{|rcll|} \hline \mathbf{\tan^2\left(\frac{x}{2}\right)} &=& \mathbf{\dfrac{1-\cos(x)}{1+\cos(x)}} \\\\ \tan^2\left(\frac{x}{2}\right) &=& \dfrac{1-\dfrac{2ab}{a^2 + b^2}}{1+\dfrac{2ab}{a^2 + b^2}} \\\\ \tan^2\left(\frac{x}{2}\right) &=& \dfrac{\dfrac{a^2 + b^2-2ab}{a^2 + b^2}}{\dfrac{a^2 + b^2+2ab}{a^2 + b^2}} \\\\ \tan^2\left(\frac{x}{2}\right) &=& \dfrac{a^2 + b^2-2ab}{a^2 + b^2+2ab} \\\\ \tan^2\left(\frac{x}{2}\right) &=& \dfrac{\left(a-b \right)^2}{\left(a+b \right)^2} \\\\ \mathbf{\tan^2\left(\frac{x}{2}\right)} &=& \mathbf{\left(\dfrac{a-b}{a+b} \right)^2} \\ \hline \end{array}\)
