+118654 I drew a right angled triangle to help me answer this question. sides 1,x and sqrt(1+x^2)
If
\(if\;\;x\ge0\;\;and\;\;\theta=tan^{-1}x\\then\\sin\theta=\ \frac{x}{\sqrt{1+x^2}}\\ cos\theta = \frac{1}{\sqrt{1+x^2}}\\ cos(tan^{-1}x)\\=cos(2\theta)\\=cos^2\theta-sin^2\theta\\ =\frac{1}{1+x^2}-\frac{x^2}{1+x^2}\\ =\frac{1-x^2}{1+x^2}\)
+26388 Write cos(2 tan−1(x)) as an algebraic expression in x.
\(\begin{array}{|rcll|} \hline && \cos(~2\cdot \underbrace{ tan^{-1}(x) }_{=\varphi}~) \qquad & | \qquad tan^{-1}(x) = \varphi \text{ or } x = \tan(\varphi) \\\\ \text{Formula:} \\ \cos(~2\cdot \varphi) &=& \dfrac{1-\tan^2(\varphi)}{1+\tan^2(\varphi)} \qquad & | \qquad \tan(\varphi) = x\\ \cos(~2\cdot \varphi) &=& \dfrac{1-x^2 }{1+x^2} \\ \mathbf{ \cos(~2\cdot \tan^{-1}(x) ) }& \mathbf{=} & \mathbf{\dfrac{1-x^2 }{1+x^2}} \\ \hline \end{array} \)
