Given sinθ=- 3/5 and cscθ=-5/3 in quadrant iii, find the value of other trigonometric functions using a pythagorean identity. show your work. find the value of cosθ, secθ and tanθ.

Let ABC be a triangle in the 3rd quadrant, right-angled at B. 
 So, AB-> Perpendicular BC -> Base AC -> Hypotenuse. 
Given: sinθ=-3/5 cosecθ=-5/3 
 According to Pythagorean theorem, square of the hypotenuse is equal to the sum of square of the other two sides.
 Therefore in triangle ABC, 〖AC〗^2=〖AB〗^2+〖BC〗^2
--(1)
 Since sinθ=Perpendicular/Hypotenuse ,
 AC=5 and AB=3
 Substituting these values in equation (1)
 
〖BC〗^2=〖AC〗^2-〖AB〗^2

 ă€–BC〗^2=5^2-3^2
 

 ă€–BC〗^2=25-9

 ă€–BC〗^2=16

 BC=4 units

 Since the triangle is in the 3rd quadrant, all trigonometric ratios, except tan
and cot are negative.
 So,cosθ=Base/Hypotenuse Cosθ=-4/5 
 secθ=Hypotnuse/Base secθ=-5/4 
 tanθ=Perpendicular/Base tanθ=3/4  
 cotθ=Base/Perpendicular cotθ=4/3

What’s the improper fraction?