Question

find \\(\sec \theta\\) if the point \\((-4, -3)\\) is on the terminal side of the \\(\theta\\) in standard po\
(a) 5/4 \quad (b) - 5/4 \quad (c) 4/5 \quad (d) -3/4 \quad (e) -5/3 \quad (f) 4/3\
\bigcirc a\
\bigcirc b\
\bigcirc c\
\bigcirc d\
\bigcirc e\
\bigcirc f\
\
question 16\
\\(\cot(-a) = \cot(a)\\)\
\bigcirc true\
\bigcirc false\
\
question 17\
\\(\sec(-a) = \sec(a)\\)\
\bigcirc true\
\bigcirc false

Explanation:

Question (about finding \(\sec\theta\))

Step1: Find the radius \(r\)

Given the point \((-4, -3)\) on the terminal side of \(\theta\), we use the formula \(r=\sqrt{x^{2}+y^{2}}\). Here, \(x = - 4\) and \(y=-3\), so \(r=\sqrt{(-4)^{2}+(-3)^{2}}=\sqrt{16 + 9}=\sqrt{25} = 5\).

Step2: Recall the definition of \(\sec\theta\)

The secant function is defined as \(\sec\theta=\frac{r}{x}\) (since \(\cos\theta=\frac{x}{r}\) and \(\sec\theta=\frac{1}{\cos\theta}\)).

Step3: Substitute the values of \(r\) and \(x\)

We know \(r = 5\) and \(x=-4\), so \(\sec\theta=\frac{5}{-4}=-\frac{5}{4}\).

The cotangent function is odd, i.e., \(\cot(-A)=-\cot(A)\). So the statement \(\cot(-A)=\cot(A)\) is false.

The secant function is even because \(\cos(-A)=\cos(A)\) (cosine is even), and \(\sec\theta=\frac{1}{\cos\theta}\). So \(\sec(-A)=\frac{1}{\cos(-A)}=\frac{1}{\cos(A)}=\sec(A)\), so the statement is true.

Answer:

b. -5/4

Question 16