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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 ?

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Well, we can use that:

$$\cos\left(\arctan\left(x\right)\right)=\frac{1}{\sqrt{1+x^2}}\tag1$$

And:

$$\sin\left(\text{arccot}\left(x\right)\right)=\frac{1}{x\cdot\sqrt{1+\frac{1}{x^2}}}\tag2$$

So in your case, we need to solve:

$$\frac{1}{\sqrt{1+x^2}}=\frac{1}{\frac{3}{4}\cdot\sqrt{1+\frac{1}{\left(\frac{3}{4}\right)^2}}}=\frac{4}{5}\space\Longleftrightarrow\space x=\pm\frac{3}{4}\tag3$$

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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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  • $\begingroup$ thnx. but wht abt considering the domains ?. does it not set any restrictions ? $\endgroup$ – ss1729 Feb 2 '18 at 7:21
  • $\begingroup$ The domain of $\arctan$ is $\mathbb R$. $\endgroup$ – Michael Rozenberg Feb 2 '18 at 7:28
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$$\cos(\arctan x)=\cos\left(\arctan\dfrac34\right)$$

$$\implies\arctan x=2m\pi\pm\arctan\dfrac34=2m\pi+\arctan\left(\pm\dfrac34\right)$$ where $m$ is any integer as $\arctan(-a)=-\arctan(a)$

As $-\dfrac\pi2<\arctan y\le\dfrac\pi2,$

$$\arctan x= \arctan\left(\pm\dfrac34\right)$$

Apply tan on both sides

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