Trigonometric Identities Question 205
Question: If A and B are acute postitive angles satisfying the equations 3 $ {{\sin }^{2}}A+2{{\sin }^{2}}B=1 $ and 3 $ \sin 2A-2\sin 2B=0 $ then A+2B is equal to
Options:
A) $ \pi $
B) $ \frac{\pi }{2} $
C) $ \frac{\pi }{4} $
D) $ \frac{\pi }{6} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ 3{{\sin }^{2}}A+2{{\sin }^{2}}B=1 $ Or $ 3{{\sin }^{2}}A=\cos 2B $ Also $ 3\sin 2A-2\sin 2B=0 $ Or $ \sin 2B=\frac{3}{2}\sin 2A $ Now, $ \cos (A+2B)=cosAcos2B-sinAsin2B $ $ =\cos A3{{\sin }^{2}}A-sinA\frac{3}{2}\sin 2A $ $ =3{{\sin }^{2}}A\cos A=3{{\sin }^{2}}AcosA=0 $
$ \therefore ,A+2B=\frac{\pi }{2} $
BETA
