Prove that: tan(A+B)/cot(A−B) = (tan²A − tan²B)/(1 − tan²A tan²B)
Question
Prove that:
\[ \frac{\tan(A+B)}{\cot(A-B)} = \frac{\tan^2 A-\tan^2 B} {1-\tan^2 A\tan^2 B} \]
Proof
L.H.S.
\[ = \frac{\tan(A+B)}{\cot(A-B)} \]
\[ = \tan(A+B)\tan(A-B) \]
\[ = \frac{\tan A+\tan B}{1-\tan A\tan B} \times \frac{\tan A-\tan B}{1+\tan A\tan B} \]
\[ = \frac{(\tan A+\tan B)(\tan A-\tan B)} {(1-\tan A\tan B)(1+\tan A\tan B)} \]
\[ = \frac{\tan^2 A-\tan^2 B} {1-\tan^2 A\tan^2 B} \]
R.H.S.
\[ = \frac{\tan^2 A-\tan^2 B} {1-\tan^2 A\tan^2 B} \]
L.H.S. = R.H.S.
Hence proved.