Question

Please help me ill mark u as brilliant

Answer 1

Answer:

see explanation

Step-by-step explanation:

Given

cotθ = $\frac{4}{5}$ = $\frac{adjacent}{opposite}$

Use Pythagoras' identity to calculate the hypotenuse h

h² = 4² + 5² = 16 + 25 = 41 ( take square root of both sides )

h = $\sqrt{41}$

θ is in quadrant III where sinθ < 0 and secθ < 0

sinθ = $\frac{opposite}{hypotenuse}$ = - $\frac{5}{\sqrt{41} }$ = -  $\frac{5\sqrt{41} }{41}$

cosθ = $\frac{adjacent}{hypotenuse}$ = - $\frac{4}{\sqrt{41} }$

secθ = $\frac{1}{cos0}$ = - $\frac{\sqrt{41} }{4}$