Trigonometric Identities Question 203
Question: $ \frac{{{\sin }^{2}}A-{{\sin }^{2}}B}{\sin A\cos A-\sin B\cos B} $ is equal to
Options:
A) $ \tan (A-B) $
B) $ \tan (A+B) $
C) $ cot(A-B) $
D) $ cot(A+B) $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ \frac{{{\sin }^{2}}A-{{\sin }^{2}}B}{\sin A\cos A-\sin B\cos B} $ $ =,\frac{2\sin (A+B)sin(A-B)}{\sin 2A-\sin 2B} $ $ =,\frac{2\sin (A+B)sin(A-B)}{2\sin (A-B)cos(A+B)} $ $ =\tan (A+B) $
BETA
