If sin(cot-'(x + 1) = cos(tan-' x), then x = (c) O (d)

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Given that ( sin left[cot ^{-1}(x+1)right]=cos left(tan ^{-1} xright) ldots(1) ) We know that, ( cot ^{-1} A=sin ^{-1} frac{1}{sqrt{1+A^{2}}} ) Here, ( A=x+1 ) Applying this identity in equation ( (1), ) we have, ( sin left[sin ^{-1} frac{1}{sqrt{1+(1+x)^{2}}}right]=cos left(tan ^{-1} xright) ldots .(2) ) A/so we know that, ( tan ^{-1} A=cos ^{-1} frac{1}{sqrt{1+A^{2}}} ) Here, ( A=x ) Applying this identity in equation ( (2), ) we have, ( sin left[sin ^{-1} frac{1}{sqrt{1+(1+x)^{2}}}right]=cos left(cos ^{-1} frac{1}{sqrt{1+x^{2}}}right) ) ( Rightarrow frac{1}{sqrt{1+(1+x)^{2}}}=frac{1}{sqrt{1+x^{2}}} ) Squaring and reciprocating both the sides, we have, ( 1+(1+x)^{2}=1+x^{2} ) ( Rightarrow 1+1+x^{2}+2 x=1+x^{2} ) ( Rightarrow 1+2 x=0 ) ( Rightarrow x=frac{-1}{2} )

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