Question

If tanA+secA=√3, 0<A<π (radians)
find A


note: All values are in radian measure​

Answer 1

secA+tanA=3–√sec⁡A+tan⁡A=3
1cosA+sinAcosA=3–√1cos⁡A+sin⁡Acos⁡A=3
1+sinA=3–√cosA1+sin⁡A=3cos⁡A
3–√cosA−sinA=13cos⁡A−sin⁡A=1
At this point we turn the coefficients (3–√,−1)(3,−1) into polar coordinates (r,θ).(r,θ). It’s in the fourth quadrant.
r=3–√2+(−1)2−−−−−−−−−−√=4–√=2r=32+(−1)2=4=2
θ=arctan2−13–√=−30∘θ=arctan2−13=−30∘
We have 3–√=2cos(−30∘)3=2cos⁡(−30∘) and −1=2sin(−30∘)−1=2sin⁡(−30∘) so we can rewrite our equation
2cos(−30)