\((m+2)\sin\theta + (2m-1)\cos\theta = 2m+1\) is correct if -
\(\tan\theta = \frac{2m}{m^2-1}\)
The equation (m+2) sin θ + (2m−1) cos θ = 2m+1 is solved using the tangent-half-angle substitution t = tan(θ/2).
Replacing sin θ = 2t/(1+t²) and cos θ = (1−t²)/(1+t²) and simplifying gives 2m t² − (m+2)t + 1 = 0.
Factoring yields t = 1/2 or t = 1/m.
Applying tan θ = 2t/(1−t²) to t = 1/m gives tan θ = 2m/(m²−1).
Hence, = 2m/m^2-1.
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