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Question

let a, b, c are non zero constant number then limrcosarcosbrcoscrsinbrsincr equals to

A
a2+b2c22bc
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B
c2+a2b22bc
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C
b2+c2a22bc
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D
independent of a, b and c
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Solution

The correct option is B b2+c2a22bc
limrcosarcosbrcoscrsinbrsincr
=limrcosar12(cos(b+cr)+cos(bcr))12(cos(bcr)cos(b+cr))
By Applying L Hospital,s Rule
=limrasinar12((b+c)sin(b+cr)(bc)sin(bcr))12(bc)sin(bcr)+12(b+c)sin(b+cr) (here1r2

wouldcanceloutfromnumeratoranddenominator)
=limra2cosar+12((b+c)2cos(b+cr)+(bc)2cos(bcr))12(bc)2cos(bcr)+12(b+c)2cos(b+cr)
=b2+c2a22bc

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