Question

Prove that: $\frac{\tan (A+B)}{\cot (A-B)}=\frac{{\tan }^{2}A-{\tan }^{2}B}{1-{\tan }^{2}A{\tan }^{2}B}$

Answer 1

Now,

$\frac{\tan (A+B)}{\cot (A-B)}$

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

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

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

$=\frac{{\tan }^{2}A-{\tan }^{2}B}{1-{\tan }^{2}A.{\tan }^{2}B}$.