Trigonometric Identities Question 368
Question: If $ \sin A=\frac{4}{5} $ and $ \cos B=-\frac{12}{13}, $ where A and B lie in first and third quadrant respectively, then $ \cos (A+B)= $
Options:
A) $ \frac{56}{65} $
B) $ -\frac{56}{65} $
C) $ \frac{16}{65} $
D) $ -\frac{16}{65} $
Show Answer
Answer:
Correct Answer: D
Solution:
We have $ \sin A=\frac{4}{5} $ and $ \cos B=-\frac{12}{13} $ Now, $ \cos ,(A+B)=\cos A,\cos B-\sin A,\sin B $ $ =\sqrt{1-\frac{16}{25}},( -\frac{12}{13} )-\frac{4}{5}\sqrt{1-\frac{144}{169}} $ $ =-\frac{3}{5}\times \frac{12}{13}-\frac{4}{5},( -\frac{5}{13} )=-\frac{16}{65} $ (Since A lies in first quadrant and B lies in third quadrant).
BETA
