Question

cscθ = \frac{1}{\sinθ} = \underline{\quad\quad} \\\\ secθ = \frac{1}{\cosθ} = \underline{\quad\quad} \\\\ cotθ = \frac{1}{\tanθ} = \underline{\quad\quad}

Explanation:

Step1: Recall reciprocal and quotient identities for Cscθ

We know that \(\csc\theta=\frac{1}{\sin\theta}\), and also from the definition of sine and the hypotenuse - opposite side in a right - triangle (or using the identity in terms of cosine and sine), but more directly, we can express \(\csc\theta\) in terms of the ratio of hypotenuse to opposite side in a right - triangle, but in terms of other trigonometric functions, we can also use the fact that \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), so \(\csc\theta = \frac{\text{hypotenuse}}{\text{opposite}}\). But in terms of other trigonometric functions, we can also use the identity \(\csc\theta=\frac{1}{\sin\theta}\), and if we want to express it in terms of cosine and cotangent, we know that \(\cot\theta=\frac{\cos\theta}{\sin\theta}\), so \(\csc\theta=\frac{\cot\theta}{\cos\theta}\) (not necessary here). The basic reciprocal identity for \(\csc\theta\) is \(\frac{1}{\sin\theta}\), but we can also express it as the ratio of the hypotenuse to the opposite side in a right - triangle (in the context of a right - triangle with angle \(\theta\)). But the standard reciprocal identity is \(\csc\theta=\frac{1}{\sin\theta}\), and also, we can express \(\csc\theta\) in terms of the sides of a right - triangle: if in a right - triangle, the length of the hypotenuse is \(r\), the length of the side opposite to \(\theta\) is \(y\), then \(\sin\theta=\frac{y}{r}\), so \(\csc\theta=\frac{r}{y}\).

Step2: Recall reciprocal and quotient identities for Secθ

We know that \(\sec\theta = \frac{1}{\cos\theta}\). In a right - triangle, if the length of the hypotenuse is \(r\) and the length of the side adjacent to \(\theta\) is \(x\), then \(\cos\theta=\frac{x}{r}\), so \(\sec\theta=\frac{r}{x}\). Also, from the quotient identities, we can relate it to other functions, but the basic reciprocal identity is \(\frac{1}{\cos\theta}\).

Step3: Recall reciprocal and quotient identities for Cotθ

We know that \(\cot\theta=\frac{1}{\tan\theta}\), and since \(\tan\theta=\frac{\sin\theta}{\cos\theta}\), then \(\cot\theta=\frac{\cos\theta}{\sin\theta}\). In a right - triangle, if the length of the side adjacent to \(\theta\) is \(x\) and the length of the side opposite to \(\theta\) is \(y\), then \(\tan\theta=\frac{y}{x}\), so \(\cot\theta=\frac{x}{y}\).

For \(\csc\theta=\frac{1}{\sin\theta}\), another way to write it (using the right - triangle definition or the ratio of hypotenuse to opposite side) is \(\csc\theta=\frac{\text{Hypotenuse}}{\text{Opposite}}\) (in a right - triangle with angle \(\theta\)). But in terms of the reciprocal identity, it's \(\frac{1}{\sin\theta}\).

For \(\sec\theta=\frac{1}{\cos\theta}\), it can also be written as \(\frac{\text{Hypotenuse}}{\text{Adjacent}}\) (in a right - triangle with angle \(\theta\)).

For \(\cot\theta=\frac{1}{\tan\theta}\), and since \(\tan\theta = \frac{\sin\theta}{\cos\theta}\), \(\cot\theta=\frac{\cos\theta}{\sin\theta}\) or \(\frac{\text{Adjacent}}{\text{Opposite}}\) (in a right - triangle with angle \(\theta\)).

If we are to fill in the blanks with the ratio of sides (for a right - triangle with angle \(\theta\), hypotenuse \(r\), opposite side \(y\), adjacent side \(x\)):

• \(\csc\theta=\frac{1}{\sin\theta}=\frac{r}{y}\)
• \(\sec\theta=\frac{1}{\cos\theta}=\frac{r}{x}\)
• \(\cot\theta=\frac{1}{\tan\theta}=\frac{x}{y}\) (or \(\frac{\cos\theta}{\sin\theta}\))

Answer:

\(\csc\theta=\frac{1}{\sin\theta}=\boldsymbol{\frac{\text{Hypotenuse}}{\text{Opposite}}}\) (or \(\frac{r}{y}\))

\(\sec\theta=\frac{1}{\cos\theta}=\boldsymbol{\frac{\text{Hypotenuse}}{\text{Adjacent}}}\) (or \(\frac{r}{x}\))

\(\cot\theta=\frac{1}{\tan\theta}=\boldsymbol{\frac{\text{Adjacent}}{\text{Opposite}}}\) (or \(\frac{\cos\theta}{\sin\theta}\) or \(\frac{x}{y}\))