Express $\cos(\tan^{-1}(x))$ in terms of $x$.
My Attempt:
Let $\tan^{-1}(x)=A$.
$$x=\tan (A)$$
Then,
$$\cos(\tan^{-1}(x))=\cos (A)$$
$\endgroup$ 33 Answers
$\begingroup$Use the relation $$1+\tan^2(A) = \frac{1}{\cos^2(A)}.$$ If you set $A=\tan^{-1}(x)$, you obtain $$1+x^2 = \frac{1}{\cos^2(\tan^{-1}(x))},$$ so that $$\cos(\tan^{-1}(x)) = \dfrac{1}{\sqrt{1+x^2}}.$$
$\endgroup$ $\begingroup$$$\tan(A) = x$$ $$\frac{\sin(A)}{\cos(A)} = x$$ $$\frac{\sin^2(A)}{\cos^2(A)} = x^2$$ $$\frac{1-\cos^2(A)}{\cos^2(A)} = x^2$$ $$\frac{1}{\cos^2(A)}-1 = x^2$$ $$\frac{1}{\cos^2(A)} = x^2+1$$ $$\cos^2(A) = \frac{1}{1+x^2}$$ $$\cos(A) = \sqrt{\frac{1}{1+x^2}}$$ $$\cos(\arctan(x)) = \sqrt{\frac{1}{1+x^2}}$$
$\endgroup$ $\begingroup$I always find it helpful to draw a picture. Note that $$-\dfrac{\pi}{2} < \arctan x < \dfrac{\pi}{2}$$
and hence $$\cos(\arctan x) > 0$$
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