# Solve inverse trigonometric equation $\sin\left(\operatorname{cot^{-1}}(x + 1)\right) = \cos\left(\tan^{-1}x\right)$

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?

• $\sin x = \sin y$ does not imply $x = y$ – ab123 Jan 15 at 16:39
• $\sin$ isn't an injective function... – Don Thousand Jan 15 at 16:44

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

$$m=0$$

Now replace $$x$$ with $$\dfrac1h$$

and consider +/- sign one by one

$$+$$ sign will give $$h=0$$

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

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 $$x=-\frac12$$.