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If \( \Large \cos \theta = \cos \alpha \cos \beta .\ then\ \tan \frac{ \theta + \alpha }{2} \tan \frac{ \theta - \alpha }{2} \) is equal to:
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A) \( \Large \tan^{2}\frac{ \alpha }{2} \) |
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B) \( \Large \tan^{2}\frac{ \beta }{2} \) |
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C) \( \Large \tan^{2}\frac{ \theta }{2} \) |
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D) \( \Large \cot^{2}\frac{ \beta }{2} \) |
Correct Answer:
| B) \( \Large \tan^{2}\frac{ \beta }{2} \) |
Description for Correct answer:
Since, \( \Large \tan \frac{ \theta + \alpha }{2}\tan \frac{ \theta \alpha }{2} = \frac{\tan^{2}\frac{ \theta }{2}-\tan^{2}\frac{ \alpha }{2}}{1-\tan^{2}\frac{ \theta }{2}\tan^{2}\frac{ \theta }{2}} \)
= \( \Large \frac{\frac{1-\cos \theta }{1+\cos \theta } - \frac{1-\cos \alpha }{1+\cos \alpha }}{1 - \frac{1-\cos \theta }{1+\cos \theta }.\frac{1-\cos \alpha }{1+\cos \alpha }} = \frac{2 \left(\cos \alpha - \cos \theta \right) }{2 \left(\cos \alpha + \cos \theta \right) } \)
= \( \Large \frac{\cos \alpha \left(1-\cos \beta \right) }{\cos \alpha \left(1+\cos \beta \right) } \)
\( \Large \left(\because \cos \theta = \cos \alpha \cos \beta \right) \)
= \( \Large \tan^{2}\frac{ \beta }{2} \)
Since, \( \Large \tan \frac{ \theta + \alpha }{2}\tan \frac{ \theta \alpha }{2} = \frac{\tan^{2}\frac{ \theta }{2}-\tan^{2}\frac{ \alpha }{2}}{1-\tan^{2}\frac{ \theta }{2}\tan^{2}\frac{ \theta }{2}} \)
= \( \Large \frac{\frac{1-\cos \theta }{1+\cos \theta } - \frac{1-\cos \alpha }{1+\cos \alpha }}{1 - \frac{1-\cos \theta }{1+\cos \theta }.\frac{1-\cos \alpha }{1+\cos \alpha }} = \frac{2 \left(\cos \alpha - \cos \theta \right) }{2 \left(\cos \alpha + \cos \theta \right) } \)
= \( \Large \frac{\cos \alpha \left(1-\cos \beta \right) }{\cos \alpha \left(1+\cos \beta \right) } \)
\( \Large \left(\because \cos \theta = \cos \alpha \cos \beta \right) \)
= \( \Large \tan^{2}\frac{ \beta }{2} \)
Part of solved Trigonometric ratio questions and answers : >> Elementary Mathematics >> Trigonometric ratio
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