# Find the values of the other trigonometric functions of theta if cot theta= -4/3and sin theta< 0.

Question
Trigonometric Functions
Find the values of the other trigonometric functions of theta if $$\displaystyle{\cot{\theta}}=-\frac{{4}}{{3}}{\quad\text{and}\quad}{\sin{\theta}}{<}{0}$$</span>.

2021-02-10
$$\displaystyle{\cot{\theta}}=−{43}$$
therefore,
$$\displaystyle{\left|{\cot{\theta}}\right|}={\left|−\frac{{4}}{{3}}\right|}=\frac{{4}}{{3}}$$
as we know that $$\displaystyle{\left|{\cot{\theta}}\right|}=$$(base)/(perpendicular)
therefore,
$$\displaystyle{\left|{\cot{\theta}}\right|}=$$(base)/(perpendicular)=4/3ZSK
therefore,
base=4 and perpendicular=3
as we know that:
$$\displaystyle{\left(\text{ypotenuse})^2=(\text{perpendicular})^2+(\text{base})^}{y}{p}{o}{t}{e}\nu{s}{e}\right\rbrace}{)}^{{2}}={\left(\text{perpendicular}\right)}^{{2}}+{\left(\text{base}\right)}^{{2}}$$
therefore,
$$\displaystyle{\left(\text{ypotenuse})^2=(\text{perpendicular})^2+(\text{base})^}{y}{p}{o}{t}{e}\nu{s}{e}\right\rbrace}{)}^{{2}}={\left(\text{perpendicular}\right)}^{{2}}+{\left(\text{base}\right)}^{{2}}$$
$$\displaystyle{\left(\text{hypotenuse}\right)}^{{2}}={\left({3}\right)}^{{2}}+{\left({4}\right)}^{{2}}$$
$$\displaystyle{\left(\text{hypotenuse}\right)}^{{2}}={9}+{16}$$
$$\displaystyle{\left(\text{hypotenuse}\right)}^{{2}}={25}$$
$$\displaystyle{\left(\text{hypotenuse}\right)}=\sqrt{{25}}$$
$$\displaystyle{\left(\text{hypotenuse}\right)}={5}$$
therefore hypotenuse=5
therefore,
$$\displaystyle{\left|{\sin{\theta}}\right|}$$=(perpendicular)/(hypotenuse)=$$\displaystyle\frac{{3}}{{5}}$$
$$\displaystyle{\left|{\cos{\theta}}\right|}$$=(base)/(hypotenuse)=$$\displaystyle\frac{{4}}{{5}}$$
$$\displaystyle{\left|{\tan{\theta}}\right|}$$=(perpendicular)/(base)=$$\displaystyle\frac{{3}}{{4}}$$
$$\displaystyle{\left|{\cos{{e}}}{c}{t}\hat{{e}}\right|}$$=(hypotenuse)/(perpendicular)=$$\displaystyle\frac{{5}}{{3}}$$
$$\displaystyle{\left|{\sec{\theta}}\right|}$$=(hypotenuse)/((base)=$$\displaystyle\frac{{5}}{{4}}$$
as $$\displaystyle{\cot{\theta}}=−{43}{\quad\text{and}\quad}{\sin{\theta}}{<}{0}$$</span>
that implies both cotangent and sine function have negative values.
as we know that:
(1) in the first quadrant all trigonometric functions are positive.
(2) in the second quadrant sine and cosecant trigonometric functions are positive and rest of trigonometric functions are negative.
(3) in the third quadrant tangent and cotangent trigonometric functions are positive and rest of trigonometric functions are negative.
(4) in the fourth quadrant cosine and secant trigonometric functions are positive and rest of trigonometric functions are negative.
as both cotangent and sine function have negative values.
therefore θ is lying in the fourth quadrant.
therefore cosine and secant trigonometric functions will be positive and rest of trigonometric functions will be negative.
therefore,
$$\displaystyle{\left|{\sin{\theta}}\right|}$$=(perpendicular)/(hypotenuse)=$$\displaystyle-\frac{{3}}{{5}}$$
$$\displaystyle{\left|{\cos{\theta}}\right|}$$=(base)/(hypotenuse)=$$\displaystyle\frac{{4}}{{5}}$$
$$\displaystyle{\left|{\tan{\theta}}\right|}$$=(perpendicular)/(base)=$$\displaystyle-\frac{{3}}{{4}}$$
$$\displaystyle{\left|{\cos{{e}}}{c}{t}\hat{{e}}\right|}$$=(hypotenuse)/(perpendicular)=$$\displaystyle-\frac{{5}}{{3}}$$
$$\displaystyle{\left|{\sec{\theta}}\right|}$$=(hypotenuse)/((base)=$$\displaystyle\frac{{5}}{{4}}$$

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