If sin θ = a²-b²/(b²-a²), find all the ratios.
Anonymous:
dunno
Let's see. Am I supposed to take a²-b² as perpendicular and b²-a² as the hypotenuse. Im learning this concepts first time so help me out.
lol. nvm.
Hi Amaan. Long time.
don't mind the theta sign.
Answers
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sin∅ = (a² - b²)/(b² - a²)
sin∅ = -(b² - a²)/(b² - a²) = -1
sin∅ = sin(-π/2)
∅ = nπ + (-1)ⁿ(-π/2)
so, smallest value of ∅ in [-π, π] is -π/2
now,
cos∅ = cos(-π/2) = 0
sec∅ = 1/0 = undefined
cosec∅ = 1/sin∅ = 1/-1 = -1
tan∅ = tan(-π/2) = undefined
cot∅ = 0
sin∅ = -(b² - a²)/(b² - a²) = -1
sin∅ = sin(-π/2)
∅ = nπ + (-1)ⁿ(-π/2)
so, smallest value of ∅ in [-π, π] is -π/2
now,
cos∅ = cos(-π/2) = 0
sec∅ = 1/0 = undefined
cosec∅ = 1/sin∅ = 1/-1 = -1
tan∅ = tan(-π/2) = undefined
cot∅ = 0
I didn't think of it that way! Thank you.
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