Trigonometric Identities Question 295
Question: If $ (1+\sin \alpha )(1+\sin \beta )(1+\sin \gamma )=(1-\sin \alpha ) $ $ (1-\sin \beta )(1-\sin \gamma )=k, $ then k is equal to:
Options:
A) $ 2\cos \alpha \cos \beta \cos \gamma $
B) $ -\cos \alpha \cos \beta \cos \gamma $
C) $ +\cos \alpha \cos \beta \cos \gamma $
D) $ +2sin\alpha sin\beta sin\gamma $
Show Answer
Answer:
Correct Answer: C
Solution:
If $ (1+\sin \alpha )(1+\sin \beta )(1+sin\gamma )=k $ And $ (1-\sin \alpha ),(1-\sin \beta ),(1-\sin \gamma )=k $ The the value of $ k^{2}=k.k. $ $ =(1+\sin \alpha )(1+sin\beta )(1+sin\gamma )(1-sin\alpha ) $ $ (1-\sin \beta )(1-sin\gamma ) $ $ =(1-{{\sin }^{2}}\alpha )(1-{{\sin }^{2}}\beta )(1-{{\sin }^{2}}\gamma ) $
$ \Rightarrow k^{2}={{\cos }^{2}}\alpha {{\cos }^{2}}\beta {{\cos }^{2}}\gamma $
$ \therefore ,k=+\cos \alpha cos\beta cos\gamma . $
BETA
