cos−1ax−cos−1bx=cos−11a−cos−11b
⇒cos−1ax+cos−11b=cos−11a+cos−1bx
We know that cos−1x+cos−1y=cos−1(xy−√1−x2√1−y2) if x,y>0 and x2+y2≤1
⇒cos−1(ax×1a−√1−a2x2√1−a2x2)=cos−1⎛⎝bx×1b−√1−b2x2√1−b2x2⎞⎠
⇒1x−√(x2−a2x2)(a2−1a2)=1x−
⎷(x2−b2x2)(b2−1b2)
⇒−√(x2−a2x2)(a2−1a2)=−
⎷(x2−b2x2)(b2−1b2)
Squaring both sides, we get
⇒(x2−a2)(a2−1)x2a2=(x2−b2)(b2−1)x2b2
⇒(x2−a2)(a2−1)a2=(x2−b2)(b2−1)b2
⇒b2(x2−a2)(a2−1)=a2(x2−b2)(b2−1)
⇒b2(x2a2−x2−a4+a2)=a2(x2b2−x2−b4+b2)
⇒b2x2a2−b2x2−b2a4+b2a2=a2x2b2−a2x2−a2b4+a2b2
⇒a2x2−b2x2=b2a4−a2b4
⇒(a2−b2)x2=a2b2(a2−b2)
⇒x2=a2b2
∴x=ab