cistG

2021-08-12

Simplify the expression:
$\mathrm{sin}\left({\mathrm{tan}}^{-1}x\right)$

### Answer & Explanation

avortarF

Solution:
Let $a={\mathrm{tan}}^{-1}x$ then from the properties of inverse trigonometry,
$\alpha ={\mathrm{tan}}^{-1}x$
$\mathrm{tan}\alpha =x$
$\mathrm{sin}\left({\mathrm{tan}}^{-1}x\right)=\mathrm{sin}\alpha$
Now we know from the definition from inverse function as $\mathrm{tan}a=x$ so it can be interpreted as,
$\mathrm{tan}\alpha =x$
$\mathrm{tan}\alpha =\frac{x}{1}=\frac{\text{perependicular}}{\text{Base}}$
Use Pythagorean formula to find he hypotenuse as,
${\left(\text{Hypotenuse}\right)}^{2}=\left({\text{Perpendicular)}}^{2}+{\left(\text{Base}\right)}^{2}$
${H}^{2}={x}^{2}+1$
$H=\sqrt{{x}^{2}+1}$
Use the above result in the formula of sin function as,
$\mathrm{sin}\left({\mathrm{tan}}^{-1}x\right)=\mathrm{sin}\alpha$
$\mathrm{sin}\alpha =\frac{\text{Perependicular}}{\text{hypotenuse}}$
$\mathrm{sin}\left({\mathrm{tan}}^{-1}x\right)=\frac{x}{\sqrt{{x}^{2}+1}}$

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