 # Simplify tan(sin^{-1}(x)) This question aims to simplify a trigonometric expression. In mathematics, trigonometric functions (also called circular functions, angle functions, or trigonometric functions) are fundamental functions that relate angle of a right triangle to the ratios of two side lengths.

They are widely used in all geometry-related sciences, such as navigation, solid mechanics, celestial mechanics, geodesy, and many others. They are among the most specific periodic functions and are also widely used to study periodic phenomena using Fourier analysis.

The trigonometric functions most used in modern mathematics are sine, cosine, and tangent. Their reciprocals are cosecant, secant, and cotangent, which are less commonly used. Each of these six trigonometric functions has a corresponding inverse function and an analog among the hyperbolic functions.

If an acute angle $\theta$ is given, then all right triangles with an angle $\theta$ are similar. This means that ratio of any two side lengths depends only on $\theta$. Therefore, these six ratios define the six functions of $\theta$, trigonometric functions.

In the following definitions, the hypotenuse is the length of the side opposite the right angle; the perpendicular represents the side opposite the given angle $\theta$, and the base represents the side between the angle $\theta$ and the right angle. $sine$

$\sin\theta=\dfrac{perpendicular}{hypotenuse}$

$cosine$

$\cos\theta=\dfrac{base}{hypotenuse}$

$tangent$

$\tan\theta=\dfrac{perpendicular}{base}$

$cosecant$

$\csc\theta=\dfrac{hypotenuse}{perpendicular}$

$secant$

$\sec\theta=\dfrac{hypotenuse}{base}$

$cotangent$

$\cot\theta=\dfrac{base}{perpendicular}$

The Pythagorean theorem is the fundamental relationship in Euclidean geometry between the three sides of a right triangle. It states that the area of a square whose side is hypotenuse (side opposite the right angle) is equal to the sum of areas of squares on the other two sides. This theorem can be stated as an equation relating lengths of the arms $a$, $b$, and hypotenuse $c$, often called the Pythagorean equation.

$c^{2}=a^{2}+b^{2}$

Let:

$\sin^{-1}(x)=\theta$

Then,

$x=\sin(\theta)$

When drawing a right-angle triangle with a hypotenuse side equal to $1$ and the other side equal to $x$.

Using the Pythagorean theorem, the third side is:

$\sqrt{1-x^{2}}$

Thus, formula for the $\tan\theta$ is given as:

$\tan\theta=\dfrac{\sin\theta}{\cos \theta}$

$=\dfrac{\sin \theta}{\sqrt{1-\sin^{2}\theta}}$

As

$x=\sin\theta$

Now we have

$\tan\theta=\dfrac{x}{\sqrt{1-x^{2}}}$

From $\sin^{-1}(x)=\theta$

We get:

$\tan(\sin^{-1}(x))=\dfrac{x}{\sqrt{1-x^{2}}}$

## Numerical Result

$\tan(\sin^{-1}(x))=\dfrac{x}{\sqrt{1-x^{2}}}$

## Example

Simplify $\cot(sin^{-1}(x))$

Let

$\sin^{-1}(x)=\theta$

Then,

$x=\sin(\theta)$

When drawing a right-angle triangle with a hypotenuse side equal to $1$ and the other side equal to $x$.

Using the Pythagorean theorem, the third side is:

$\sqrt{1-x^{2}}$

Thus, formula for the $cot\theta$ is given as:

$\cot\theta=\dfrac{\cos\theta}{\sin \theta}$

$=\dfrac{\sqrt{1-\sin^{2}\theta}}{\sin \theta}$

As

$x=\sin\theta$

Now we have:

$\cot\theta=\dfrac{\sqrt{1-x^{2}}}{x}$

From $\sin^{-1}(x)=\theta$

We get:

$\cot(\sin^{-1}(x))=\dfrac{\sqrt{1-x^{2}}}{x}$