Trigonometric Identities Question 224
Question: If $ \theta $ and $ \varphi $ are angles in the 1st quadrant such that $ \tan \theta =1/7 $ and $ \sin \varphi =1/\sqrt{10} $ .Then
[Kurukshetra CEE 1998; AMU 2001]
Options:
A) $ \theta +2\varphi =90{}^\circ $
B) $ \theta +2\varphi =60{}^\circ $
C) $ \theta +2\varphi =30{}^\circ $
D) $ \theta +2\varphi =45{}^\circ $
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Answer:
Correct Answer: D
Solution:
Given, $ \tan \theta =\frac{1}{7},\sin \varphi =\frac{1}{\sqrt{10}} $ $ \sin \theta =\frac{1}{\sqrt{50}},\cos \theta =\frac{7}{\sqrt{50}},\cos \varphi =\frac{3}{\sqrt{10}} $
$ \therefore \cos 2\varphi =2{{\cos }^{2}}\varphi -1=2.\frac{9}{10}-1=\frac{8}{10} $ $ \sin 2\varphi =2\sin \varphi \cos \varphi =2\times .\frac{1}{\sqrt{10}}\times \frac{3}{\sqrt{10}}=\frac{6}{10} $ \ $ \cos (\theta +2\varphi )=\cos \theta \cos 2\varphi -\sin \theta \sin 2\varphi $ $ =\frac{7}{\sqrt{50}}\times \frac{8}{10}-\frac{1}{\sqrt{50}}.\frac{6}{10} $ $ =\frac{56-6}{10\sqrt{50}}=\frac{50}{10\sqrt{50}}=\frac{5\sqrt{2}}{10}=\frac{1}{\sqrt{2}} $ \ $ \theta +2\varphi =45^{o} $ .
BETA
