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Question

Find the value of x if,
cos1axcos1bx=cos11bcos11a

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Solution

cos1axcos1bx=cos11acos11b

cos1ax+cos11b=cos11a+cos1bx

We know that cos1x+cos1y=cos1(xy1x21y2) if x,y>0 and x2+y21

cos1(ax×1a1a2x21a2x2)=cos1bx×1b1b2x21b2x2

1x(x2a2x2)(a21a2)=1x (x2b2x2)(b21b2)

(x2a2x2)(a21a2)= (x2b2x2)(b21b2)

Squaring both sides, we get

(x2a2)(a21)x2a2=(x2b2)(b21)x2b2

(x2a2)(a21)a2=(x2b2)(b21)b2

b2(x2a2)(a21)=a2(x2b2)(b21)

b2(x2a2x2a4+a2)=a2(x2b2x2b4+b2)

b2x2a2b2x2b2a4+b2a2=a2x2b2a2x2a2b4+a2b2

a2x2b2x2=b2a4a2b4

(a2b2)x2=a2b2(a2b2)

x2=a2b2

x=ab

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