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
〖BC〗^2=5^2-3^2
〖BC〗^2=25-9
〖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
