Trigonometric Identities Question 246
Question: If A lies in the third quadrant and $ 3,\tan A-4=0, $ then $ 5,\sin 2A+3,\sin A+4,\cos A= $
[EAMCET 1994]
Options:
A) 0
B) $ \frac{-24}{5} $
C) $ \frac{24}{5} $
D) $ \frac{48}{5} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ 3\tan A-4=0\Rightarrow \tan A=\frac{4}{3}\Rightarrow \sin A=-\frac{4}{5},\cos A=-\frac{3}{5} $
$ \therefore $ $ 5\sin 2A+3\sin A+4\cos A $ = $ 10\sin A\cos A+3\sin A+4\cos A $ = $ 10,( \frac{12}{25} )-\frac{12}{5}-\frac{12}{5}=0 $ .
BETA
