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If $\sin\left(\operatorname{cot^{-1}}(x + 1)\right) = \cos\left(\tan^{-1}x\right)$, then find the value of $x$.

Please solve this question by using $\cos\left(\dfrac\pi2 - \theta\right) = \sin\theta$ by changing $\cos\left(\tan^{-1}x\right) = \sin\left(\dfrac\pi2 - \tan^{-1}x\right)$ and then equate both LHS and RHS. If not then why? How does the contradiction below occur?

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    $\begingroup$ $\sin x = \sin y$ does not imply $x = y$ $\endgroup$ – ab123 Jan 15 at 16:39
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    $\begingroup$ $\sin$ isn't an injective function... $\endgroup$ – Don Thousand Jan 15 at 16:44
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Replace $x$ with $\dfrac1h$ to find $h=0$

Alternatively

$$\cos(\arctan x)=\sin(\text{arccot}(x+1))=\cdots=\cos(\arctan(x+1))$$

$$\arctan(x)=2m\pi\pm\arctan(x+1)$$ where $m$ is an integer

Using https://en.m.wikipedia.org/wiki/Inverse_trigonometric_functions#Principal_values

$m=0$

Now replace $x$ with $\dfrac1h$

and consider +/- sign one by one

$+$ sign will give $h=0$

By using $-$ sign, $h=-2$

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Note that there are multiple possibilities to the equation you derived below,

$$\sin(\cot^{-1}(1+x))=\sin(\frac\pi2-\tan^{-1} x)$$

You only considered

$$\cot^{-1}(1+x) = \frac\pi2-\tan^{-1} x$$ which leads to contradiction, or, no solutions. In addition, you also need to examine

$$\cot^{-1}(1+x) = \pi - (\frac\pi2-\tan^{-1} x)$$

which leads to $x+1=-x$, hence the valid solution $x=-\frac12$.

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