$mg=2T\sin 45°$

$\Rightarrow mg=\sqrt{2}T$

  $T_1\text{ cos}\theta =Tcos45^{\circ }$

$\Rightarrow T_1\text{cos}\theta =\frac{T}{\sqrt{2}}=\frac{\mathit{\text{mg}}}{2}$              ...(i)

Further, $\mathit{\text{Mg}}+T\text{ cos }45^{\circ }={\mathit{\text{T}}}_1\text{ sin}\mathrm{\theta }\left\{\because \mathit{\text{T}}=\frac{\mathit{\text{mg}}}{\sqrt{2}}\right\}$

$\Rightarrow T_1\text{ sin}\theta =\mathit{\text{Mg}}+\frac{\mathit{\text{mg}}}{\sqrt{2}}\frac{1}{\sqrt{2}}$              ...(ii)

$\Rightarrow \text{tan}\theta =\frac{\mathit{\text{Mg}}+\frac{\mathit{\text{mg}}}{2}}{\frac{\mathit{\text{mg}}}{2}}=1+\frac{2\mathit{\text{M}}}{\mathit{\text{m}}}$

 {dividing Eq. (ii) by Eq. (i)}