Trigonometric Identities Question 57
Question: If $ \cos (\theta -\alpha ),\ \cos \theta $ and $ \cos (\theta +\alpha ) $ are in H.P., then $ \cos \theta \sec \frac{\alpha }{2} $ is equal to
[IIT 1997]
Options:
A) $ \pm \sqrt{2} $
B) $ \pm \sqrt{3} $
C) $ \pm 1/\sqrt{2} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Given $ \cos (\theta -\alpha ),\cos \theta $ and $ \cos (\theta +\alpha ) $ are in H.P.
Þ $ \frac{1}{\cos (\theta -\alpha )},\frac{1}{\cos \theta },\frac{1}{\cos (\theta +\alpha )} $ will be in A.P. Hence, $ \frac{2}{\cos \theta }=\frac{1}{\cos (\theta -\alpha )}+\frac{1}{\cos (\theta +\alpha )} $ $ =\frac{\cos (\alpha +\theta )+\cos (\theta -\alpha )}{{{\cos }^{2}}\theta -{{\sin }^{2}}\alpha } $
Þ $ \frac{2}{\cos \theta }=\frac{2\cos \theta \cos \alpha }{{{\cos }^{2}}\theta -{{\sin }^{2}}\alpha } $
Þ $ {{\cos }^{2}}\theta -{{\sin }^{2}}\alpha ={{\cos }^{2}}\theta \cos \alpha $
Þ $ {{\cos }^{2}}\theta ,(1-\cos \alpha )={{\sin }^{2}}\alpha $
Þ $ {{\cos }^{2}}\theta ( 2{{\sin }^{2}}\frac{\alpha }{2} )=4{{\sin }^{2}}\frac{\alpha }{2}{{\cos }^{2}}\frac{\alpha }{2} $ $ {{\cos }^{2}}\theta {{\sec }^{2}}\frac{\alpha }{2}=2\Rightarrow \cos \theta \sec \frac{\alpha }{2}=\pm \sqrt{2} $ .
BETA
