If $\alpha $, $\beta $, $\gamma $, $\delta $ are smallest positive angle in ascending order of magnitude which have their sines equal to positive quantity $\lambda $ , then if

$4\mathrm{sin}\frac{\alpha}{2}+3\mathrm{sin}\frac{\beta}{2}+2\mathrm{sin}\frac{\gamma}{2}+\mathrm{sin}\frac{\delta}{2}=2\sqrt{1+\lambda}$

then $\lambda $ = ?

Options-

(1)$\mathrm{sin}\alpha $

(2)$\mathrm{sin}\beta $

(3)$\mathrm{sin}\gamma $

(4)$\mathrm{sin}\delta $

$4\mathrm{sin}\frac{\alpha}{2}+3\mathrm{sin}\frac{\beta}{2}+2\mathrm{sin}\frac{\gamma}{2}+\mathrm{sin}\frac{\delta}{2}=2\sqrt{1+\lambda}$

then $\lambda $ = ?

Options-

(1)$\mathrm{sin}\alpha $

(2)$\mathrm{sin}\beta $

(3)$\mathrm{sin}\gamma $

(4)$\mathrm{sin}\delta $