Question

Sec theta = 5/3 and the terminal point determined by quadrant 4

Answer 1

Answer:

Here, we have $\sec\theta =\frac{5}{3}$

Since $\sec\theta=\frac{Hypotenuse}{Base}= \frac{5}{3}$

So, for the angle θ,

Hypotenuse = 5 and Base = 3

Using Pythogoras theorem,

(Hypotenuse)² = (Base²) + (Perpendicular )²

(5 )² = ( 3 )² + (Perpendicular )²

⇒ Perpendicular = 25 - 9 = √16 = 4

we have given θ lie in IV quadrant. In IV quadrant cos and sec are positive function but sine, cosec, tan and cot are negative.

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

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

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

$\csc\theta=\frac{Hypotenuse}{Perpendicular}=-\frac{5}{4}$

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

Answer 2

Cot theta = -3/4
-
sin theta = -4/5

Note - The - is just a spacer your not subtracting.