Express $\cos(\tan^{-1}(x))$ in terms of $x$.

Question

Express $\cos(\tan^{-1}(x))$ in terms of $x$.

My Attempt:

Let $\tan^{-1}(x)=A$.

$$x=\tan (A)$$

Then,

$$\cos(\tan^{-1}(x))=\cos (A)$$

Answer 1

$$\tan(A) = x$$ $$\frac{\sin(A)}{\cos(A)} = x$$ $$\frac{\sin^2(A)}{\cos^2(A)} = x^2$$ $$\frac{1-\cos^2(A)}{\cos^2(A)} = x^2$$ $$\frac{1}{\cos^2(A)}-1 = x^2$$ $$\frac{1}{\cos^2(A)} = x^2+1$$ $$\cos^2(A) = \frac{1}{1+x^2}$$ $$\cos(A) = \sqrt{\frac{1}{1+x^2}}$$ $$\cos(\arctan(x)) = \sqrt{\frac{1}{1+x^2}}$$

Answer 2

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 $$\cos(\tan^{-1}(x)) = \dfrac{1}{\sqrt{1+x^2}}.$$

Answer 3

I always find it helpful to draw a picture. Note that $$-\dfrac{\pi}{2} < \arctan x < \dfrac{\pi}{2}$$

and hence $$\cos(\arctan x) > 0$$