If secB=x,tanB=y(cosecA-cotA)=4/3,sinA+cosA=?
cscA-cotA = (1-cosA)/sinA = tan(A/2) = 4/3
so, tanA = 2tanA/(1 - tan^2A) = 4/3
3tanA = 2 - 2tan^2A
2tan^2A + 3tanA - 2 = 0
tanA = (-3±5)/4
tanA = -2 or 1/2
Now, knowing tanA, you can find sinA and cosA
Not sure what B has to do with anything there...
To find sinA + cosA, we need to first find the values of sinA and cosA individually. Let's start by using the given information to find those values.
We are given that secB = x and tanB = y(cosecA - cotA) = 4/3. We can use these values to find sinB and cosB by using the following formulas:
1. secB = 1/cosB
Therefore, cosB = 1/secB = 1/x
2. tanB = sinB/cosB
Substituting the values of tanB and cosB, we get:
y(cosecA - cotA) = sinB / cosB
y(cosecA - cotA) = sinB / (1/x)
y(cosecA - cotA) = x * sinB
sinB = y(cosecA - cotA) / x
Now that we have the values of sinB and cosB, we can use the Pythagorean identity sin^2B + cos^2B = 1 to find sinA and cosA.
sinA = sinB / cosB
cosA = √(1 - sin^2A)
By substituting sinB and cosB into the above formulas, we could find the values of sinA and cosA.