Trigonometric Identities Question 372
Question: $ \frac{{{\sin }^{2}}A-{{\sin }^{2}}B}{\sin A\cos A-\sin B\cos B}= $
[MP PET 1993]
Options:
A) $ \tan (A-B) $
B) $ \tan (A+B) $
C) $ \cot (A-B) $
D) $ \cot (A+B) $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \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,\cos ,(A+B),\sin ,(A-B)}=\tan ,(A+B) $ .
BETA
