Let θ₁ and θ₂ denote the angles made by the tension forces T₁ and T₂ with the positive horizontal axis (directly to the right), respectively. (BTW, I write T₁ in boldface for the vector and T₁ for its magnitude.)
The ball is kept in equilibrium, so by Newton's second law both the net horizontal and net vertical forces are zero. This means
[1] T₁ cos(θ₁) + T₂ cos(θ₂) = 0
[2] T₁ sin(θ₁) + T₂ sin(θ₂) - m g = 0
Both cos(x) and sin(x) fall within -1 and 1 for any angle x, so right away we know that
T₁ sin(θ₁) + T₂ sin(θ₂) ≠ T₁ + T₂
so option (B) is not correct.
If T₁ = T₂, then in equation [1] we get
T₁ cos(θ₁) + T₁ cos(θ₂) = 0
T₁ (cos(θ₁) + cos(θ₂)) = 0
cos(θ₁) + cos(θ₂) = 0
which means θ₁ = 180° - θ₂, since cos(180° - x) = - cos(x) for any x. But this would require both strings to have the same length. So option (C) is also not correct.
The horizontal components of T₁ and T₂ are the only forces acting in the horizontal direction, so their horizontal components are equal. However, the vertical component of T₂ is closer to being parallel to the weight of the ball, which means T₂'s vertical component is almost enough on its own to counteract the ball's weight. This in turn means that (D)T₂ > T₁.
In other words, sin(θ₂) is closer to 1 than sin(θ₁), and T₂ is closer to m g than T₁.
