If cot−1(1−x22x)+cos−1(1−x21+x2)=2π3,x>0 , x≠1 then x=
A
1√2
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B
±1√2
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C
±1√3
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D
1√3
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Solution
The correct option is D1√3 cot−1(1−x22x)=2π3−cos−1(1−x21+x2) cos−1(1−x21+x2)=2π3−cot−1(1−x22x) take cos on both sides Simplify, 1−x21+x2=−12(1−x21+x2)+√322x1+x2[cos(A−B)=cosAcosB+sinAsinB] 2−2x2−2√3x2(1+x2)=−12(1−x2)(1+x2) 2−2x2−2√3x=x2−1 3x2+2√3x−3=0 ∴x=−2√3±√12+366 x=−2√3±4√36 x=−√3,√33 x=−√3,1√3