if 2sinx=1,π÷2<x<π and √2cosy=1,
3π÷2<y<2π, find the value of
tanx + tany÷ cosx - cosy
To find the value of tanx + tany ÷ cosx - cosy, we first need to find the values of x and y using the given equations.
From the equation 2sinx = 1, we can solve for x by dividing both sides of the equation by 2. This gives us sinx = 1/2.
Since π/2 < x < π, we are looking for the quadrant where sin is positive and cosine is negative. This corresponds to the second quadrant (where the sine is positive and the cosine is negative).
In the second quadrant, the sine function is positive, but the cosine function is negative. In this case, we can use the Pythagorean identity to determine the value of sin(x). The Pythagorean identity states that sin^2(x) + cos^2(x) = 1.
Since we know that cos(x) is negative in the second quadrant, we can use the positive value of sin(x) to find its value as sin^2(x) + cos^2(x) = 1.
Substituting sin^2(x) = 1/4, we get 1/4 + cos^2(x) = 1.
Rearranging the equation, we have cos^2(x) = 3/4.
Taking the square root of both sides, we find that cos(x) = -√3/2.
Similarly, from the equation √2cos(y) = 1, we can solve for y by dividing both sides of the equation by √2. This gives us cos(y) = 1/√2.
Since 3π/2 < y < 2π, we are looking for the quadrant where cosine is positive and sine is negative. This corresponds to the fourth quadrant (where both sine and cosine are negative).
In the fourth quadrant, both sine and cosine are negative. In this case, we can use the Pythagorean identity and the known value of cos(y) to find the value of sin(y).
Using sin^2(y) + cos^2(y) = 1, and cos(y) = 1/√2, we can solve for sin(y):
sin^2(y) + (1/√2)^2 = 1
sin^2(y) + 1/2 = 1
sin^2(y) = 1 - 1/2 = 1/2
Taking the square root of both sides, we find that sin(y) = -1/√2.
Now that we have the values of sin(x) and cos(x) in the second quadrant, and sin(y) and cos(y) in the fourth quadrant, we can calculate the value of tan(x) and tan(y).
Recall that tan(x) = sin(x) / cos(x). Therefore, tan(x) = (1/2) / (-√3/2) = -1/√3 = -√3/3.
Similarly, tan(y) = sin(y) / cos(y). Therefore, tan(y) = (-1/√2) / (1/√2) = -1.
Finally, we can substitute these values into the expression tanx + tany ÷ cosx - cosy:
tanx + tany ÷ cosx - cosy = (-√3/3) + (-1) ÷ (-√3/2) - (-1/√2)
= -√3/3 - 2/√3 + 1/√2
To combine the fractions, we need to find a common denominator. The common denominator is √6, which is the product of the denominators of √3 and √2.
Therefore, the expression can be written as:
= (-√3/3) (√6/√6) - (2/√3)(√2/√2) + (1/√2)(√3/√3)
= -√(18)/3 + (-2√6)/√(6) + √3/√2
= -√18/3 - 2√6/√6 + √3/√2
Now, let's simplify the expression further by rationalizing the denominators:
= -√18/3 - 2√6/√6 + √3/√2 * √2/√2
= -√18/3 - 2√6/√6 + √6/√4
= -√18/3 - 2√6/√6 + √6/2
= -√18/3 - √6/√6 + √6/2
= -√18/3 - 1 + √6/2
To calculate further, we need to find a common denominator for the fractions. The common denominator is 6:
= (-√18 - 6 + 3√6)/6
Thus, the value of tanx + tany ÷ cosx - cosy is (-√18 - 6 + 3√6)/6.
if 2sinx =1 ,π/2<x<π also √2 ×cosy=1, 3π/2<y<2π, find the value of tanx+tany÷cosx-cosy
sinx = 1/2
cosx = -√3/2
siny = -1/√2
cosy = 1/√2
Now just plug in your values:
(-1/√3 - 1)/(-√3/2 - 1/√2) =
I'll let you massage that to a more pleasant appearance.