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 ?