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If \( \Large \cos \left( \theta - \alpha \right),\ \cos \theta\ and\ \cos \left( \theta + \alpha \right) \) are in HP, the: \( \Large \cos \theta \sec \frac{ \alpha }{2} \) is equal to:
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A) \( \Large \pm \sqrt{2} \) |
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B) \( \Large \pm \sqrt{3} \) |
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C) \( \Large \pm \frac{1}{\sqrt{2}} \) |
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D) none of these |
Correct Answer:
| A) \( \Large \pm \sqrt{2} \) |
Description for Correct answer:
Given \( \Large \cos \left( \theta - \alpha \right) \cos \theta\ and\ \cos \left( \theta + \alpha \right)\ \) are in HP
=> \( \Large \frac{1}{\cos \left( \theta - \alpha \right) },\ \frac{1}{\cos \theta },\ \frac{1}{\cos \left( \theta + \alpha \right) } \) will be in AP.
=> \( \Large \frac{2}{\cos \theta }=\frac{1}{\cos \left( \theta - \alpha \right) }+\frac{1}{\cos \left( \theta + \alpha \right) } \)
= \( \Large \frac{\cos \left( \alpha + \theta \right)+\cos \left( \theta - \alpha \right) }{\cos^{2} \theta - \sin^{2} \alpha } \)
=> \( \Large \frac{2}{\cos \theta } = \frac{2 \cos \theta \cos \alpha }{\cos^{2} \theta - \sin^{2} \alpha } \)
=> \( \Large \cos^{2} \theta - \sin^{2} \alpha = \cos^{2} \theta \cos \alpha \)
=> \( \Large \cos^{2} \theta \left(1-\cos \alpha \right) = \sin^{2} \alpha \)
=> \( \Large \cos^{2} \theta \left(2 \sin^{2}\frac{ \alpha }{2}\right) = 4 \sin^{2}\frac{ \alpha }{2} \cos^{2}\frac{ \alpha }{2} \)
=> \( \Large \cos^{2} \theta \sec^{2}\frac{ \alpha }{2} = 2 \)
=> \( \Large \cos \theta \sec^{2}\frac{ \alpha }{2} = \pm \sqrt{2} \)
Given \( \Large \cos \left( \theta - \alpha \right) \cos \theta\ and\ \cos \left( \theta + \alpha \right)\ \) are in HP
=> \( \Large \frac{1}{\cos \left( \theta - \alpha \right) },\ \frac{1}{\cos \theta },\ \frac{1}{\cos \left( \theta + \alpha \right) } \) will be in AP.
=> \( \Large \frac{2}{\cos \theta }=\frac{1}{\cos \left( \theta - \alpha \right) }+\frac{1}{\cos \left( \theta + \alpha \right) } \)
= \( \Large \frac{\cos \left( \alpha + \theta \right)+\cos \left( \theta - \alpha \right) }{\cos^{2} \theta - \sin^{2} \alpha } \)
=> \( \Large \frac{2}{\cos \theta } = \frac{2 \cos \theta \cos \alpha }{\cos^{2} \theta - \sin^{2} \alpha } \)
=> \( \Large \cos^{2} \theta - \sin^{2} \alpha = \cos^{2} \theta \cos \alpha \)
=> \( \Large \cos^{2} \theta \left(1-\cos \alpha \right) = \sin^{2} \alpha \)
=> \( \Large \cos^{2} \theta \left(2 \sin^{2}\frac{ \alpha }{2}\right) = 4 \sin^{2}\frac{ \alpha }{2} \cos^{2}\frac{ \alpha }{2} \)
=> \( \Large \cos^{2} \theta \sec^{2}\frac{ \alpha }{2} = 2 \)
=> \( \Large \cos \theta \sec^{2}\frac{ \alpha }{2} = \pm \sqrt{2} \)
Part of solved Trigonometric ratio questions and answers : >> Elementary Mathematics >> Trigonometric ratio
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