Question
If $$\left( {1 + \sin \alpha } \right)\left( {1 + \sin \beta } \right)\left( {1 + \sin \gamma } \right) = \left( {1 - \sin \alpha } \right)\left( {1 - \sin \beta } \right)\left( {1 - \sin \gamma } \right) = k,$$ then $$k$$ is equal to :
A.
$$2 \cos \alpha \cos \beta \cos \gamma $$
B.
$$ - \cos \alpha \cos \beta \cos \gamma $$
C.
$$ + \cos \alpha \cos \beta \cos \gamma $$
D.
$$ + 2 \sin \alpha \sin \beta \sin \gamma $$
Answer :
$$ + \cos \alpha \cos \beta \cos \gamma $$
Solution :
If $$\left( {1 + \sin \alpha } \right)\left( {1 + \sin \beta } \right)\left( {1 + \sin \gamma } \right) = k$$
And $$\left( {1 - \sin \alpha } \right)\left( {1 - \sin \beta } \right)\left( {1 - \sin \gamma } \right) = k$$
The value of $${k^2} = k \cdot k.$$
$$\eqalign{
& = \left( {1 + \sin \alpha } \right)\left( {1 + \sin \beta } \right)\left( {1 + \sin \gamma } \right)\left( {1 - \sin \alpha } \right)\left( {1 - \sin \beta } \right)\left( {1 - \sin \gamma } \right) \cr
& = \left( {1 + \sin \alpha } \right)\left( {1 - \sin \alpha } \right)\left( {1 + \sin \beta } \right)\left( {1 - \sin \beta } \right)\left( {1 + \sin \gamma } \right)\left( {1 - \sin \gamma } \right) \cr
& = \left( {1 - {{\sin }^2}\alpha } \right)\left( {1 - {{\sin }^2}\beta } \right)\left( {1 - {{\sin }^2}\gamma } \right) \cr
& \Rightarrow {k^2} = {\cos ^2} \alpha \,{\cos ^2} \beta \,{\cos ^2} \gamma \cr
& \therefore k = + \cos \alpha \cos \beta \cos \gamma . \cr} $$