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.**

\[c^{2}=a^{2}+b^{2}\]

**Expert Answer**

**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:**

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