Question

Write the given expression as an algebraic expression in x. tan(2 cos^-1(x))

Answer 1

$\bf tan\left[ 2cos^{-1}(x) \right]\implies tan\left[ 2cos^{-1}\left( \frac{x}{1} \right) \right] \\\\\\ \textit{if we say }cos^{-1}\left( \frac{x}{1} \right)=\theta\textit{ that means }tan\left[ 2cos^{-1}\left( \frac{x}{1} \right) \right]\iff tan(2\theta)\\\\ -----------------------------\\\\ cos^{-1}\left( \frac{x}{1} \right)=\theta\implies cos(\theta)=\cfrac{x}{1}\cfrac{\leftarrow adjacent=a}{\leftarrow hypotenuse=c}\\\\ -----------------------------$

$\bf \textit{again, using the pythagorean theorem to get the opposite side} \\\\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\implies \pm\sqrt{1^2-x^2}=b \\\\\\ \pm\sqrt{1-x^2}=b\\\\ -----------------------------\\\\ tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\implies tan(2\theta)=\cfrac{2\cdot \frac{\pm\sqrt{1-x^2}}{x}}{1-\left( \frac{\pm\sqrt{1-x^2}}{x} \right)^2}$

$\bf tan(2\theta)=\cfrac{\frac{\pm2\sqrt{1-x^2}}{x}}{1-\frac{1-x}{x}}\implies tan(2\theta)=\cfrac{\frac{\pm2\sqrt{1-x^2}}{x}}{\frac{x-1+x}{x}} \\\\\\ tan(2\theta)=\cfrac{\pm2\sqrt{1-x^2}}{x}\cdot \cfrac{x}{2x-1} \implies tan(2\theta)=\cfrac{\pm 2\sqrt{1-x^2}}{2x-1}$

Answer 2

Answer: The algebraic expression for the given expression is $\frac{\pm 2x^{2}\sqrt{1-x^2} }{2x^{2}-1}$.

Step-by-step explanation:

Let us assume $cos^{-1} x= y$ ⇒ $cosy=x=\frac{x}{1}$

⇒$tany=\pm \frac{\sqrt{1-{x^{2}}} }{x}$

$[cosy=\frac{x}{1} =\frac{base}{hypotenuse} ,which\: means, tany = \frac{perpendicular}{base} =\frac{hypotenuse^2-base^2}{base} ]$

Now,

$x.tan(2cos^{-1}x)= x.tan(2y)$

$=x.\frac{2tany}{1-tan^{2}y}$

$=x.\frac{\pm 2\sqrt{1-x^{2}}/x }{1-\frac{1-x^{2}}{x^2} }$

$=\frac{\pm 2 x^{2}\sqrt{1-x^2} }{2x^{2}-1}$

Thus, $x.tan(2cos^{-1}x)=$$\frac{\pm 2 x^{2}\sqrt{1-x^2} }{2x^{2}-1}$

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