Question

If $x=\tan (A+B)+\tan (A-B),y=\tan (A+B)\tan (A-B)$ then

• A. $x=\frac{2\sin 2A}{\cos 2A+\cos 2B}$
• B. $x=\frac{\sin 2A}{{\cos }^{2}A+{\cos }^{2}B}$
• C. $y=\frac{\cos 2B-\cos 2A}{\cos 2B+\cos 2A}$
• D. $y=\frac{\cos 2A-\cos 2B}{\cos 2A+\cos 2B}$

Answer 1

Answer: (A), (C)

$x=\frac{2\sin 2A}{\cos 2A+\cos 2B}$
$y=\frac{\cos 2B-\cos 2A}{\cos 2B+\cos 2A}$

$x=\tan (A+B)+\tan (A-B)$

$=\frac{\tan A+\tan B}{1-\tan A\tan B}+\frac{\tan A-\tan B}{1+\tan A\tan B}$

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

$=\frac{\sin (A+B)}{\cos (A+B)}+\frac{\sin (A-B)}{\cos (A-B)}$

$=\frac{\sin (A+B)\cos (A-B)+\sin (A-B)\cos (A+B)}{\cos (A+B)\cos (A-B)}$

$=\frac{\sin 2A}{\cos 2A+\cos 2B}$

And

$y=\tan (A+B)\times \tan (A-B)$

$=\frac{\tan A+\tan B}{1-\tan A\tan B}\times \frac{\tan A-\tan B}{1+\tan A\tan B}$

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

$=\frac{\sin (A+B)}{\cos (A+B)}\times \frac{\sin (A-B)}{\cos (A-B)}=\frac{\cos 2B-\cos 2A}{\cos 2B+\cos 2A}$