- $cot\theta$
- $sin\theta$
**Where**$\theta$**in Quadrant II**

This problem aims to familiarize us with **trigonometric functions.** The concepts required to solve this problem are related to **trigonometry,** which includes **quadrantal** **angles** and **signs** of **function.**

The **sign** of a **trigonometric function** such as $sin\theta$ relies on the signs of the **x,y** **coordinate** points of the **angle.** We can also figure out the signs of all the **trigonometric** functions by understanding in which **quadrant** the angle lies. The terminal angle might lie in any of the **eight** regions, **4** of which are the quadrants and along the **4** axis. Each **position** represents something **additional** for the signs of the trigonometric functions.

To understand the **signs** of the **trigonometric** functions, we must understand the sign of $x$ and $y$ **coordinates.** For this, we know that **distance** between any point and origin is forever **positive,** but $x$ and $y$ can be positive or negative.

## Expert Answer

Let’s first see the **quadrants,** in the $1^{st}$ quadrant, $x$ and $y$ are all **positive,** and all $6$ **trigonometric** functions will have **positive** values. In the $2^{nd}$ quadrant, only $sin\theta$ and $cosec\theta$ are **positive.** In the $3^{rd}$ quadrant, only $tan\theta$ and $cot\theta$ are **positive.** Ultimately, in the $4^{th}$ quadrant, only $cos\theta$, and $sec\theta$ are **positive.**

Now let’s start our **solution** since $cot\theta$ is the **reciprocal** of $tan\theta$, which is **equal** to $\dfrac{$sin\theta$}{ $cos\theta$}$, so:

\[cot\theta = \dfrac{cos\theta}{sin\theta}\]

To **rewrite** $cot\theta$ only in **terms** of $sin\theta$, we have to change $cos\theta$ into $sin\theta$, using the **trigonometric identity:**

\[cos^2 \theta + sin^2 \theta = 1\]

\[cos^2 \theta = 1 – sin^2 \theta\]

\[cos\theta = \pm \sqrt{1 – sin^2 \theta}\]

Since $cos\theta$ lies in the $2^{nd}$ **quadrant,** we will apply the **negative** sign to equal its effect:

\[cot\theta = \dfrac{-cos\theta}{sin\theta}\]

\[cot\theta = \dfrac{- \sqrt{1 – sin^2 \theta}}{sin\theta}\]

Hence, this our **final expression** of $cot\theta$ in terms of $sin\theta$.

## Numerical Result

The **final expression** of $cot\theta$ in **terms** of $sin\theta$ is $\dfrac{- \sqrt{1 – sin^2 \theta} }{sin\theta}$.

## Example

Write $tan\theta$ in **terms** of the $cos\theta$, where $\theta$ lies in the $4$ **Quadrant.** Also write other **trigonometric values** in **Quad III** for $sec\theta = -2$.

**Part a:**

Since $tan\theta$ is the **fraction** of $sin\theta$ over $cos\theta$, so:

\[tan\theta=\dfrac{sin\theta}{cos\theta}\]

To write in **terms** of $cos\theta$, applying the change using the **trigonomteric identity:**

\[cos^2 \theta + sin^2 \theta = 1 \]

\[sin^2 \theta = 1 – cos^2 \theta \]

\[sin\theta = \pm \sqrt{1 – cos^2 \theta} \]

Since $sin\theta$ lies in the $4^{th}$ **quadrant,** apply **negative** sign :

\[tan\theta = \dfrac{-sin\theta}{cos\theta} \]

\[tan\theta = \dfrac{-\sqrt{1 – cos^2 \theta}}{cos\theta} \]

**Part b:**

Using the **definition** of $secant$:

\[sec\theta = \dfrac{hypotenuse}{base}\]

To find the other sides of the **right triangle** we will use the **Pythagorean** theorem:

\[H^2 = B^2 + P^2 \]

\[P = \sqrt{B^2 – H^2}\]

Since $sec$ lies in the **III Quad,** we will apply the **negative** sign:

\[ P = -\sqrt{2^2 + 1^2}\]

\[ P = -\sqrt{3}\]

Now **find** the other values:

\[ sin\theta = -\dfrac{\sqrt{3}}{2}\]

\[ cos\theta = -\dfrac{1}{2}\]

\[ tan\theta = \sqrt{3}\]

\[ cot\theta = \dfrac{\sqrt{3}}{3}\]

\[ cosc\theta = -\dfrac{2\sqrt{3}}{3}\]