1708\hspace{-16}\;$(2)\;\;Let\;$\mathbf{y=\cos x.\left(\sin x+\sqrt{\sin^2 x+\sin ^2 \alpha}\right)}$\\\\\\ $\mathbf{y=\left(\sin x.\cos x+\cos x.\sqrt{\sin^2 x+\sin^2 \alpha}\right)}$\\\\\\ Now Using $\mathbf{C.S}$ Inequality, We Get\\\\\\ $\mathbf{\left(\sin ^2 x+\cos^2 x\right).\left(\cos^2 x+\left(\sqrt{\sin^2 x+\sin^2 \alpha}\right)^2\right)\geq \left(\sin x.\cos x+\cos x.\sqrt{\sin^2 x+\sin^2 \alpha}\right)^2}$\\\\\\ $\mathbf{\left|\sin x.\cos x+\cos x.\sqrt{\sin^2 x+\sin^2 \alpha}\right|\leq \sqrt{1+\sin^2 \alpha}}$\\\\\\ So $\mathbf{-\sqrt{1+\sin^2\alpha}\left\leq (\sin x.\cos x+\cos x.\sqrt{\sin^2 x+\sin^2 \alpha}\right)\leq +\sqrt{1+\sin^2 \alpha}}$\\\\\\ So $\mathbf{-\sqrt{1+\sin^2 \alpha}\leq y\leq \sqrt{1+\sin^2 \alpha}}$\\\\\\ $\mathbf{y\in \left[-\sqrt{1+\sin^2 \alpha}\;, \sqrt{1+\sin^2 \alpha}\right]}$
