\(\sin \alpha+\cos \alpha=b \Rightarrow(\sin \alpha+\cos \alpha)^{2} =b^{2} \)
\(\sin ^{2} \alpha+\cos ^{2} \alpha+2 \sin \alpha \cos \alpha =b^{2} \)
\(\sin 2 \alpha =b^{2}-1 \)
Since \(-1<\sin 2 \alpha \leq 1 \)
\(-1 \leq \mathrm{b}^{2}-1 \leq 1 \)
\(\mathrm{~b}^{2}-1 \leq 1 \)
\(\mathrm{~b}^{2} \leq 2 \)
\(\text { This is possible if } \mathrm{b} \leq \sqrt{2}\)
\(\sin 2 \alpha=b^{2}-1, \text { if } b \leq \sqrt{2}\)

\(x^{2}+a x+b=0\)
\(\text { Let } \tan \alpha \text { and } \tan \beta \text { be the roots. }\)
\(\tan \alpha+\tan \beta =-a \)
\(\tan \alpha \tan \beta =b \)
\(\frac{\sin (\alpha+\beta)}{\sin \alpha \sin \beta} =\frac{\sin \alpha \cos \beta+\cos \alpha \sin \beta}{\sin \alpha \sin \beta} \)
\(=\cot \beta+\cot \alpha \)
\(=\frac{1}{\tan \alpha}+\frac{1}{\tan \beta} \)
\(=\frac{\tan \alpha+\tan \beta}{\tan \alpha \tan \beta}=\frac{-a}{b} \)

\(\cos 1^{\circ}+\cos 2^{\circ}+\cos 3^{\circ}+\ldots \ldots \ldots+\cos 179^{\circ} \)
\(=\left(\cos 1^{\circ}+\cos 179^{\circ}\right)+\left(\cos 2^{\circ}+\cos 178^{\circ}\right)+ ...\)
\(=2 \cos 90^{\circ} \cos 89^{\circ}+2 \cos 90^{\circ} \cos 80^{\circ}+\ldots \ldots . \)
\(=0+0+0 \ldots \ldots=0 \)

\(4 \sin ^{2} x+3 \cos ^{2} x+\sin \frac{x}{2}+\cos \frac{x}{2} \)
\(=3\left(\sin ^{2} x+\cos ^{2} x\right)+\sin ^{2} x+\sin \frac{x}{2}+\cos \frac{x}{2} \)
\(=3+\sin ^{2} x+\sin \frac{x}{2}+\cos \frac{x}{2} \)
\(\text { Maximum of } \sin x=1\)
\(\text { Maximum value of } \sin ^{2} x=1\)
\(\text { Maximum value of } \sin \frac{x}{2} \text { occurs where } x=45^{\circ}\)
\(\text { Maximum value is } 3+1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=4+\frac{2}{\sqrt{2}}=4+\sqrt{2}\)

\(x =\frac{1}{\cos 80^{\circ}}-\frac{\sqrt{3}}{\sin 80^{\circ}} \)
\(=\frac{\sin 80^{\circ}-\sqrt{3} \cos 80^{\circ}}{\sin 80^{\circ} \cos 80^{\circ}} \)
\(\frac{x}{2} =\frac{\frac{1}{2} \sin 80^{\circ}-\frac{\sqrt{3}}{2} \cos 80^{\circ}}{\frac{1}{2} \sin 80^{\circ} \cos 80^{\circ}} \)
\(=\frac{\sin 80^{\circ} \cos 60^{\circ}-\cos 80^{\circ} \sin 60^{\circ}}{\frac{1}{2} \sin 80^{\circ} \cos 80^{\circ}} \)
\(=\frac{\sin 20^{\circ}}{\frac{1}{2} \sin 80^{\circ} \cos 80^{\circ}} \)
\(=\frac{\sin 160^{\circ}}{\frac{1}{2} \sin 80^{\circ} \cos 80^{\circ}} \)
\(=\frac{2 \sin 80^{\circ} \cos 80^{\circ}}{\frac{1}{2} \sin 80^{\circ} \cos 80^{\circ}}=4 \)