91Ó°ÊÓ

Start by writing the double angle formulas for sine and cosine: \(\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\) and \(\cos \theta = 1 - 2 \sin^2 \frac{\theta}{2}\)

Recall that \(\tan \theta = \frac{\sin \theta}{\cos \theta}\). Substitute the expressions for \(\sin \theta\) and \(\cos \theta\) from Step 1: \(\tan \theta = \frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{1 - 2 \sin^2 \frac{\theta}{2}}\)

Now we have to rewrite the expression for \(\tan \theta\) to represent \(\tan \frac{\theta}{2}\) in terms of \(\tan \theta\). Recall that \(\tan \frac{\theta}{2} = \frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}\). Divide the numerator and denominator of \(\tan \theta\) by \(\cos^2 \frac{\theta}{2}\): \(\tan \theta = \frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \div \cos^2 \frac{\theta}{2}}{(1 - 2 \sin^2 \frac{\theta}{2}) \div \cos^2 \frac{\theta}{2}}\) \(\tan \theta = \frac{2 \frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}}{\frac{1}{\cos^2 \frac{\theta}{2}} - 2 \frac{\sin^2 \frac{\theta}{2}}{\cos^2 \frac{\theta}{2}}}\) Now, notice that \(\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}} = \tan \frac{\theta}{2}\) and \(\frac{\sin^2 \frac{\theta}{2}}{\cos^2 \frac{\theta}{2}} = \tan^2 \frac{\theta}{2}\): \(\tan \theta = \frac{2 \tan \frac{\theta}{2}}{\frac{1}{\cos^2 \frac{\theta}{2}} - 2 \tan^2 \frac{\theta}{2}}\)

Now, solve for \(\tan \frac{\theta}{2}\) in the equation: \(\tan \frac{\theta}{2} = \frac{\tan \theta}{2 - 2 \tan^2 \theta}\) This is the desired formula that expresses \(\tan \frac{\theta}{2}\) in terms of \(\tan \theta\).