If sin(θ)=−2/5, and θ is in quadrant III , then find cos(θ)= tan(θ)= sec(θ)= csc(θ)= cot(θ)=

Answer 1

If sin(θ) = -2/5 and θ is in quadrant III, then we can use the values of the trigonometric functions for angles in the third quadrant to find the values of the other functions.

In quadrant III, the cosine function is negative, so we have cos(θ) = -sqrt(1 - sin^2(θ)) = -sqrt(1 - (-2/5)^2) = -sqrt(1 - 4/25) = -sqrt(21/25) = -3/5.

In quadrant III, the tangent function is also negative, so we have tan(θ) = sin(θ) / cos(θ) = (-2/5) / (-3/5) = 2/3.

In quadrant III, the secant function is also negative, so we have sec(θ) = 1 / cos(θ) = 1 / (-3/5) = -5/3.

In quadrant III, the cosecant function is positive, so we have csc(θ) = 1 / sin(θ) = 1 / (-2/5) = -5/2.

In quadrant III, the cotangent function is also positive, so we have cot(θ) = 1 / tan(θ) = 1 / (2/3) = 3/2.

Therefore, the values of the trigonometric functions for θ in quadrant III are:

cos(θ) = -3/5 tan(θ) = 2/3 sec(θ) = -5/3 csc(θ) = -5/2 cot(θ) = 3/2

Answer 2

sinT= -2/5 = opposite side over hypotenuse= o/h

o=-2, h=5,

a =-sqr(h^2-o^2) = -sqr(25-4)= -sqr21

cosT = adjacent side over hypotenuse = a/h = -sqr21/5

secT = -5/sqr21 = -5sqr21/21

cscT = -5/2 = -2.5

tanT = sinT/cosT = -2/-sqr21 = 2/sqr21 = 2sqr21/21

cotT = 1/tanT = sqr21/2