MCQ
If $\sin ({\cot ^{ - 1}}(x + 1) = \cos ({\tan ^{ - 1}}x)$, then $ x =$
- ✓$ - \frac{1}{2}$
- B$\frac{1}{2}$
- C$0$
- D$\frac{9}{4}$
$ = \frac{1}{{\sqrt {{x^2} + 2x + 2} }}$
$\cos ({\tan ^{ - 1}}x) = \cos \left( {{{\cos }^{ - 1}}\frac{1}{{\sqrt {1 + {x^2}} }}} \right) = \frac{1}{{\sqrt {1 + {x^2}} }}$
Thus, $\frac{1}{{\sqrt {{x^2} + 2x + 2} }} = \frac{1}{{\sqrt {1 + {x^2}} }}$
$ \Rightarrow {x^2} + 2x + 2 = 1 + {x^2}$
$ \Rightarrow $ $x = - \frac{1}{2}$.
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