Question

If $\tan \theta =\frac{p^2-q^2}{2pq}$, then

• A. $\tan \left(\frac{\theta }{2}\right)=\frac{p-q}{p+q}$
• B. $\tan \left(\frac{\theta }{2}\right)=\frac{p+q}{p-q}$
• C. $\cot \left(\frac{\theta }{2}\right)=\frac{p-q}{p+q}$
• D. $\cot \left(\frac{\theta }{2}\right)=\frac{p+q}{p-q}$

Answer 1

Answer: (A), (D)

The correct options are
A $\tan \left(\frac{\theta }{2}\right)=\frac{p-q}{p+q}$
D $\cot \left(\frac{\theta }{2}\right)=\frac{p+q}{p-q}$

$\tan \theta =\frac{p^2-q^2}{2pq}\Rightarrow \frac{2\tan \frac{\theta }{2}}{1-{\tan }^2\frac{\theta }{2}}=\frac{p^2-q^2}{2pq}\Rightarrow 4pq\tan \frac{\theta }{2}=(p^2-q^2)-(p^2-q^2){\tan }^2\frac{\theta }{2}\Rightarrow (p^2-q^2){\tan }^2\frac{\theta }{2}+4pq\tan \frac{\theta }{2}-(p^2-q^2)\Rightarrow \tan \frac{\theta }{2}=\frac{-4pq\pm \sqrt{16p^2{q}^2+4{(p^2-q^2)}^2}}{2(p^2-q^2)}=\frac{-4pq\pm 2(p^2+q^2)}{(p^2-q^2)}=\frac{{(p-q)}^2}{(p^2-q^2)}\frac{-{(p+q)}^2}{(p^2-q^2)}$
$=\frac{p-q}{p+q}$ or $\frac{-(p+q)}{p-q}$
And $\Rightarrow \cot \frac{\theta }{2}=\frac{1}{\tan \frac{\theta }{2}}=\frac{p+q}{p-q}$ or $\frac{p-q}{-(p+q)}$