Question

If $\tan \left(\beta \right)=\cos \left(\theta \right)\tan \left(\alpha \right)$ then ${\tan }^2\left(\frac{\theta }{2}\right)=?$

• A. $\frac{\sin \left(\alpha +\beta \right)}{\sin \left(\alpha -\beta \right)}$
• B. $\frac{\cos \left(\alpha -\beta \right)}{\cos \left(\alpha +\beta \right)}$
• C. $\frac{\sin \left(\alpha -\beta \right)}{\sin \left(\alpha +\beta \right)}$
• D. $\frac{\cos \left(\alpha +\beta \right)}{\cos \left(\alpha -\beta \right)}$

Answer 1

The correct option is C

$\frac{\sin \left(\alpha -\beta \right)}{\sin \left(\alpha +\beta \right)}$

Explanation for the correct option

Given that $\tan \left(\beta \right)=\cos \left(\theta \right)\tan \left(\alpha \right)$

We know that ${\tan }^2\left(\frac{\theta }{2}\right)=\frac{{\sin }^2\left(\frac{\theta }{2}\right)}{{\cos }^2\left(\frac{\theta }{2}\right)}$

Multiplying numerator and denominator by $2$, we get,

$\begin{array}{crcll}& {\tan }^2\left(\frac{\theta }{2}\right)& =& \frac{2{\sin }^2\left(\frac{\theta }{2}\right)}{2{\cos }^2\left(\frac{\theta }{2}\right)}& \\ & & =& \frac{1-\cos \left(\theta \right)}{1+\cos \left(\theta \right)}& \left[\because 2{\sin }^2\left(A\right)=1-\cos \left(2A\right);2{\cos }^2\left(A\right)=1+\cos \left(2A\right)\right]\\ & & =& \frac{1-{\displaystyle \frac{\tan \left(\beta \right)}{\tan \left(\alpha \right)}}}{1+\frac{\tan \left(\beta \right)}{\tan \left(\alpha \right)}}& \left[\because \cos \left(\theta \right)=\frac{\tan \left(\beta \right)}{\tan \left(\alpha \right)}\right]\\ & & =& \frac{\tan \left(\alpha \right)-\tan \left(\beta \right)}{\tan \left(\alpha \right)+\tan \left(\beta \right)}& \\ & & =& \frac{{\displaystyle \frac{\sin \left(\alpha \right)}{\cos \left(\alpha \right)}}-{\displaystyle \frac{\sin \left(\beta \right)}{\cos \left(\beta \right)}}}{\frac{\sin \left(\alpha \right)}{\cos \left(\alpha \right)}+\frac{\sin \left(\beta \right)}{\cos \left(\beta \right)}}& \left[\because \tan \left(\theta \right)=\frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}\right]\\ & & =& \frac{\sin \left(\alpha \right)\cos \left(\beta \right)-\cos \left(\alpha \right)\sin \left(\beta \right)}{\sin \left(\alpha \right)\cos \left(\beta \right)+\cos \left(\alpha \right)\sin \left(\beta \right)}& \\ \Rightarrow & \begin{array}{cr}& {\tan }^2\left(\frac{\theta }{2}\right)\end{array}& =& \frac{\sin \left(\alpha -\beta \right)}{\sin \left(\alpha +\beta \right)}& \end{array}$

Hence the correct option is option(C) i.e. $\frac{\sin \left(\alpha -\beta \right)}{\sin \left(\alpha +\beta \right)}$