Trigonometric Identities Question 152
Question: If A and B are positive acute angles satisfying $ 3{{\cos }^{2}}A+2{{\cos }^{2}}B=4 $ and $ \frac{3\sin A}{\sin B}=\frac{2\cos B}{\cos A} $ Then the value of $ A+2B $ is equal to:
Options:
A) $ \frac{\pi }{6} $
B) $ \frac{\pi }{2} $
C) $ \frac{\pi }{3} $
D) $ \frac{\pi }{4} $
Show Answer
Answer:
Correct Answer: B
Solution:
Given, $ 3{{\cos }^{2}}A+2{{\cos }^{2}}B=4 $
$ \Rightarrow 2{{\cos }^{2}}B-1=4-3{{\cos }^{2}}A-1 $
$ \Rightarrow \cos 2B=3(1-{{\cos }^{2}}A)=3{{\sin }^{2}}A $ ?.(1) and $ 2\cos B,\sin B=3\sin A,\cos A $ $ \sin 2B=3\sin A\cos A $ ?..(2) Now, $ \cos (A+2B)=\cos A\cos 2B-\sin A\sin 2B $ $ =\cos A(3{{\sin }^{2}}A)-\sin A(3sin,Acos,A)=0 $ [using eqs.(1) and (2)]
$ \Rightarrow ,A+2B=\frac{\pi }{2} $
BETA
