Sec A is solved to be [tex]\frac{8}{\sqrt{55} }[/tex]
Solving cosx / secx - cotx / tan x gives cos^2 x - cot^2 x
What is quadrant of a circle?
This is division of circle into 4 parts. each part is termed a quadrant
How to solve for sec A
sine θ = opposite / hypotenuse
In the third quadrant:
The opposite = -3
hypotenuse = 8
sec θ = 1 / cos θ
cosθ = adjacent / hypotenuse
From Pythagoras theorem
hypotenuse² = adjacent² + opposite²
adjacent = √64 - 9
= √55
cos θ = √55/8
sec ∅ = 1 / cos ∅ = 8/√55
How to solve cosx / secx - cotx / tan x
from trigonometry we know that:
secx = 1 / cosx and cotx = 1 / tanx
plugging the trigonometry values to the given expression gives
cosx / secx - cotx / tan x = cosx / 1/cosx - 1/tanx / tanx
= cosx / 1/cosx - 1/tanx / tanx
= cosx * cosx - cotx * cotx
= cos^2 x - cot^2 x
Read more on quadrant here: https://brainly.com/question/25038683
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