If (1+sinx)(1+siny)(1+sinz)=(1−sinx)(1−siny)(1−sinz)=k then k has the value
A
±cosxcosycosz
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B
±sinxsinysinz
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C
±3sinxsinysinz
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D
None of these
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Solution
The correct option is A±cosxcosycosz Multiply both sides by (1+sinx)(1+siny)(1+sinz) ⇒(1+sinx)(1+siny)(1+sinz) =(1−sin2x)(1−sin2y)(1−sin2z) =cos2xcos2ycos2z ⇒(1+sinx)(1+siny)(1+sinz) =±cosxcosycosz