<s>[INST] Solving $\cos\big(\tan^{-1}x\big)=\sin\big(\cot^{-1}\frac{3}{4}\big)$ Solve

$$
\cos\big(\tan^{-1}x\big)=\sin\big(\cot^{-1}\frac{3}{4}\big)
$$

My Attempt:
From the domain consideration,
$$
\boxed{0\leq\tan^{-1}\frac{3}{4},\tan^{-1}x\leq\frac{\pi}{2}}
$$
$$
\cos\big(\tan^{-1}x\big)=\cos\big(\frac{\pi}{2}-\cot^{-1}\frac{3}{4}\big)\implies\cos\big(\tan^{-1}x\big)=\cos\big(\tan^{-1}\frac{3}{4}\big)\\\implies\tan^{-1}x=2n\pi\pm\tan^{-1}\frac{3}{4}\\
\implies \tan^{-1}x=\tan^{-1}\frac{3}{4}\quad\text{ as }0\leq\tan^{-1}\frac{3}{4}\leq\frac{\pi}{2}\\
\implies x=\frac{3}{4}
$$
Is it correct or $\frac{-3}{4}$ also is a solutions ?
What about the condition $0\leq\tan^{-1}\frac{3}{4},\tan^{-1}x\leq\frac{\pi}{2}$, does this affect the solutions ?
 [/INST] Yes because $\cos$ is an even function.
It follows also from your way:
$$\arctan{x}=\pm\arctan\frac{3}{4},$$ which gives $x=\frac{3}{4}$ or $x=-\frac{3}{4}.$ 
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