Trigonometric Identities Question 45
Question: If $ (\sec \alpha +\tan \alpha )(\sec \beta +\tan \beta )(\sec \gamma +\tan \gamma ) $ $ =\tan \alpha \tan \beta \tan \gamma $ , then $ (\sec \alpha -\tan \alpha )(\sec \beta -\tan \beta ) $ $ (\sec \gamma -\tan \gamma )= $
[Kurukshetra CEE 1998]
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:
Given: $ (\sec \alpha +\tan \alpha )(\sec \beta +\tan \beta )(\sec \gamma +\tan \gamma ) $ $ =\tan \alpha \tan \beta \tan \gamma $ …(i) Let $ x=(\sec \alpha -\tan \alpha )(\sec \beta -\tan \beta )(\sec \gamma -\tan \gamma ) $ …(ii) Multiply both equations, (i) and (ii), we get $ ({{\sec }^{2}}\alpha -{{\tan }^{2}}\alpha )({{\sec }^{2}}\beta -{{\tan }^{2}}\beta )({{\sec }^{2}}\gamma -{{\tan }^{2}}\gamma ) $ $ =x.(\tan \alpha \tan \beta \tan \gamma ) $
$ \Rightarrow x=\frac{1}{\tan \alpha \tan \beta \tan \gamma } $
$ \therefore x=\cot \alpha \cot \beta \cot \gamma $
BETA
