Trigonometric Identities Question 247
Question: If $ (\sec \alpha +\tan \alpha )(\sec \beta +\tan \beta )(\sec \gamma +\tan \gamma ) $ $ =\tan ,\alpha \tan \beta \tan \gamma , $ then expression $ (\sec \alpha -\tan \alpha ),(sec\beta -tan\beta )(sec\gamma -tan\gamma ) $ is equal to
Options:
A) $ \cot \alpha \cot \beta \cot \gamma $
B) $ \tan \alpha tan\beta tan\gamma $
C) $ \cot \alpha +\cot \beta +\cot \gamma $
D) $ tan\alpha +tan\beta +tan\gamma $
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Answer:
Correct Answer: A
Solution:
$ (\sec \alpha +\tan \alpha )(\sec \beta +\tan \beta )(\sec \gamma +\tan \gamma ) $ $ =\tan \alpha \tan \beta \tan \gamma $
$ \Rightarrow ,({{\sec }^{2}}\alpha -{{\tan }^{2}}\alpha )({{\sec }^{2}}\beta -{{\tan }^{2}}\beta )({{\sec }^{2}}\gamma -{{\tan }^{2}}\gamma ) $ $ =\tan \alpha tan\beta tan\gamma (sec\alpha -tan\alpha )(sec\beta -tan\beta ) $ $ (\sec \gamma -\tan \gamma ) $
$ \Rightarrow (\sec \alpha -\tan \alpha )(\sec \beta -\tan \beta )(\sec \gamma -\tan \gamma ) $ $ =\cot \alpha \cot \beta \cot \gamma $
BETA
