solve it...........;.;..........
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Hi ,
LHS = [(a+1/b)^m(a-1/b)^n]/[(b+1/a)^m(b-1/a)^n ]
= [ ( ab + 1 )^m/b^m × ( ab - 1 )^n/b^n ]/[ ( ab+1)^m/a^m × ( ab-1)^n/a^n ]
After cancellation of numerators ,
=[ ( 1/b^m ) ( 1/b^n ) ]/[ ( 1/a^m )( 1/a^n )]
= ( a^m × a^n )/ ( b^m × b^n )
= ( a^m+n )/( b^m+n )
= ( a/b )^m+n
= RHS
Hence proved.
I hope this helps you.
: )
LHS = [(a+1/b)^m(a-1/b)^n]/[(b+1/a)^m(b-1/a)^n ]
= [ ( ab + 1 )^m/b^m × ( ab - 1 )^n/b^n ]/[ ( ab+1)^m/a^m × ( ab-1)^n/a^n ]
After cancellation of numerators ,
=[ ( 1/b^m ) ( 1/b^n ) ]/[ ( 1/a^m )( 1/a^n )]
= ( a^m × a^n )/ ( b^m × b^n )
= ( a^m+n )/( b^m+n )
= ( a/b )^m+n
= RHS
Hence proved.
I hope this helps you.
: )
abhi569:
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