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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)$\mathrm{sin}\delta$
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percolarse2rzd
See we have ascending order for positive sines as $a,\pi -a,2\pi +a,3\pi -a$ as sine is positive only in first two quadrants .Now these are values of $\alpha ,\beta ,\gamma ,\delta$ plugging in and simplifying we get $\mathrm{sin}\left(\alpha /2\right)+\mathrm{cos}\left(\alpha /2\right)=\sqrt{1+\lambda }$ thus squaring we get $\lambda =\mathrm{sin}\left(\alpha \right)$