We have: tan(theta) + sec(theta) = 1
Rightarrow frac(sin(theta))(cos(theta)) + frac(1)(cos(theta)) = 1
Rightarrow frac(sin(theta) + 1)(cos(theta)) = 1
Rightarrow sin(theta) + 1 = cos(theta)
Rightarrow sin(theta) - cos(theta) = - 1
Then, let's consider sin(theta) - cos(theta) = R cos(theta + alpha).
We can expand the right-hand side using the compound angle identity for cos(theta):
Rightarrow sin(theta) - cos(theta) = R (cos(theta) cos(alpha) - sin(theta) sin(alpha))
Rightarrow sin(theta) - cos(theta) = R cos(theta) cos(alpha) - R sin(theta) sin(alpha)
Rightarrow sin(theta) - cos(theta) = R cos(alpha) cos(theta) - R sin(alpha) sin(theta)
If we compare the coefficients of sin(theta) and cos(theta), we get:
Rightarrow 1 = - R sin(alpha) Rightarrow R sin(alpha) = - 1 " " " " " " (i)
and
Rightarrow - 1 = R cos(alpha) Rightarrow R cos(alpha) = - 1 " " " " " " (ii)
If we divide (i) by (ii), we get:
Rightarrow frac(R sin(alpha))(R cos(alpha)) = frac(- 1)(- 1)
Rightarrow tan(alpha) = 1
Rightarrow alpha = frac(pi)(4)
If we sum the squares of (i) and (ii), we get:
Rightarrow R^(2) sin^(2)(alpha) + R^(2) cos^(2)(alpha) = 1 + 1
Rightarrow R^(2) (cos^(2)(alpha) + sin^(2)(alpha)) = 2
Rightarrow R^(2) = 2
Rightarrow R = sqrt(2)
So, we can write sin(theta) - cos(theta) in the form sqrt(2) cos(theta + frac(pi)(4)), i.e. in our original problem we get sqrt(2) cos(theta + frac(pi)(4)) = - 1:
Rightarrow cos(theta + frac(pi)(4)) = - frac(1)(sqrt(2)) = - frac(sqrt(2))(2)
Now, let the reference angle be cos(theta) = frac(sqrt(2))(2) Rightarrow theta = frac(pi)(4).
But the value of cos(theta + frac(pi)(4)) is negative, so theta must be located in either the second or third quadrant:
Rightarrow theta + frac(pi)(4) = pi - frac(pi)(4), pi + frac(pi)(4)
Rightarrow theta + frac(pi)(4) = frac(3 pi)(4), frac(5 pi)(4)
Rightarrow theta = frac(pi)(2), pi
As you have not specified a domain that theta must be located within, we can give the general solutions to the equation as:
therefore theta = frac(pi)(2) pm 2 pi n, pi pm 2 pi n; n in ZZ