Answer:
Here, we have [tex]\sec\theta =\frac{5}{3}[/tex]
Since [tex]\sec\theta=\frac{Hypotenuse}{Base}= \frac{5}{3}[/tex]
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.
[tex]\sin\theta=\frac{Perpendicular}{Hypotenuse}=-\frac{4}{5}[/tex]
[tex]\cos\theta=\frac{Base}{Hypotenuse}=\frac{3}{5}[/tex]
[tex]\tan\theta=\frac{Perpendicular}{Base}=-\frac{4}{3}[/tex]
[tex]\csc\theta=\frac{Hypotenuse}{Perpendicular}=-\frac{5}{4}[/tex]
[tex]\cot\theta=\frac{Base}{Perpendicular}=-\frac{3}{4}[/tex]
