Question

6) Find all trigonometric functions of angle in
standard position whose terminal arm passes
through point (3,-4).​

Answer 1

Answer:

Step-by-step explanation:

Let

$x=3$ and $y=-4$

$r=$ the length of a line segment drawn from the origin to the point.

$r=\sqrt{x^{2}+y^{2}  }$

⇒$r=\sqrt{3^{2}+(-4)^{2}  }$

⇒$r=\sqrt{25}$

⇒$r=5$

$sin\alpha =\frac{y}{r} =\frac{-4}{5}$

$cos\alpha =\frac{x}{r} =\frac{3}{5}$

$tan\alpha =\frac{y}{x} =\frac{-4}{3}$

$cot\alpha =\frac{x}{y} =\frac{3}{-4}$

$sec\alpha =\frac{r}{x} =\frac{5}{3}$

$cosec\alpha =\frac{r}{y} =\frac{5}{-4}$

Answer 2

Answer with Step-by-step explanation:

Given point :(3,-4)

x=3 and y=-4

$r=\sqrt{x^2+y^2}$

Using the formula

$r=\sqrt{3^2+(-4)^2}$

$r=\sqrt{9+16}=5$

We know that

$sin\theta=\frac{y}{r}, cos\theta=\frac{x}{r}$

Using the formula

$sin\theta=\frac{-4}{5}=-\frac{4}{5}$

$cos\theta=\frac{3}{5}$

$sec\theta=\frac{1}{cos\theta}=\frac{1}{\frac{3}{5}}=\frac{5}{3}$

$cosec\theta=\frac{1}{sin\theta}=-\frac{5}{4}$

$tan\theta=\frac{sin\theta}{cos\theta}$

$tan\theta=\frac{-\frac{4}{5}}{\frac{3}{5}}$

$tan\theta=-\frac{4}{3}$

$cot\theta=\frac{1}{tan\theta}$

$cot\theta=-\frac{3}{4}$

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