If cscØ = 4/3, find sinØ + sinØ cot^2 Ø
csc A = 4/3. sin A = 3/4,
x^2 + 9 = 16, X = sqrt7 = 7^(1/2).
Cot A = sqrt7/3.
sin A + sin A * cot^2 A = 3/4 + 3/4*(7^(1/2))^2 = 3/4+3/4*7 = 24/4 = 6.
To find the value of sinØ + sinØ cot²Ø given that cscØ = 4/3, we can use the reciprocal identity cscØ = 1/sinØ.
Since cscØ = 4/3, we know that 1/sinØ = 4/3. To solve for sinØ, we can take the reciprocal of both sides:
1 / (1/sinØ) = 3/4
This simplifies to:
sinØ = 3/4
Now, we can substitute this value into the expression sinØ + sinØ cot²Ø:
sinØ + sinØ cot²Ø = (3/4) + (3/4) * cot²Ø
To proceed, we need to find the value of cot²Ø. To do this, we can use the identity cot²Ø = (cosØ/sinØ)².
We already know that sinØ = 3/4, so we need to find the value of cosØ. We can use the Pythagorean Identity to do this: sin²Ø + cos²Ø = 1. Since sinØ = 3/4, we can substitute this value into the equation:
(3/4)² + cos²Ø = 1
9/16 + cos²Ø = 1
cos²Ø = 1 - 9/16
cos²Ø = 7/16
Taking the square root of both sides, we have:
cosØ = √(7/16)
Since cscØ = 4/3, we know that sinØ = 3/4, and cosØ = √(7/16).
Now, we can substitute these values into the expression sinØ + sinØ cot²Ø:
sinØ + sinØ cot²Ø = (3/4) + (3/4) * (√(7/16)/3/4)²
simplifycot²Ø to:
sinØ + sinØ cot²Ø = (3/4) + (3/4) * (√(7/16))^2
Now, simplify the expression:
sinØ + sinØ cot²Ø = (3/4) + (3/4) * (7/16)
sinØ + sinØ cot²Ø = (3/4) + (3/4) * (7/16)
sinØ + sinØ cot²Ø = 3/4 + (21/64)
sinØ + sinØ cot²Ø = (12/16) + (21/64)
sinØ + sinØ cot²Ø = 48/64 + 21/64
sinØ + sinØ cot²Ø = (48 + 21)/64
sinØ + sinØ cot²Ø = 69/64