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  • Given that: (1+cosx)(1+cosy)(1+cosz)=(1-cosx)(1-cosy)(1-cosz) . Show that one of the value of each member of equality is sinx .siny .sinz

Given that: (1+cosx)(1+cosy)(1+cosz)=(1-cosx)(1-cosy)(1-cosz) . Show that one of the value of each member of equality is sinx .siny .sinz

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Solution: We have,

(1+cosx)(1+cosy)(1+cosz)=(1-cosx)(1-cosy)(1-cosz)

Multiplying both sides by (1+cosx)(1+cosy)(1+cosz) we get

(1+cosx)^2(1+cosy)^2(1+cosz)^2=(1-cos^2x)(1-cos^2y)(1-cos^2z)\ \Rightarrow (1+cosx)^2(1+cosy)^2(1+cosz)^2 =sin^2xcdot sin^2ycdot sin^2z\ \Rightarrow (1+cosx)(1+cosy)(1+cosz)=pm sinxcdot sinycdot sinz

Hence , one of the value of (1+cosx)(1+cosy)(1+cosz) is sinxcdot sinycdot sinz

 

Posted by

Deependra Verma

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