Question

If $\cos {\theta }_1=2\cos {\theta }_2,$ then $\tan \frac{{\theta }_1-{\theta }_2}{2}$. $\tan \frac{{\theta }_1+{\theta }_2}{2}$ is equal to

• A. $\frac{1}{3}$
• B. $\frac{-1}{3}$
• C. $1$
• D. $-1$

Answer 1

The correct option is B $\frac{-1}{3}$

$\sin \frac{A-B}{2}\times \sin \frac{A+B}{2}={\sin }^2\frac{A}{2}-{\sin }^2\frac{B}{2}=1-\cos A-1+\cos B=\cos B-\cos A$

$\cos \frac{A-B}{2}\times \cos \frac{A+B}{2}={\cos }^2\frac{B}{2}-{\sin }^2\frac{A}{2}=1+\cos B-1+\cos A=\cos B+\cos A$

$\tan \frac{{\theta }_1-{\theta }_2}{2}\tan \frac{{\theta }_1+{\theta }_2}{2}=\frac{\sin \frac{{\theta }_1-{\theta }_2}{2}\sin \frac{{\theta }_1+{\theta }_2}{2}}{\cos \frac{{\theta }_1-{\theta }_2}{2}\cos \frac{{\theta }_1+{\theta }_2}{2}}=\frac{\cos {\theta }_2-\cos {\theta }_1}{\cos {\theta }_2+\cos {\theta }_1}=\frac{\cos {\theta }_2-2\cos {\theta }_2}{\cos {\theta }_2+2\cos {\theta }_2}=\frac{-\cos {\theta }_2}{3\cos {\theta }_2}=\frac{-1}{3}$