If cosα + cosβ = 3 2 and sinα + sinβ = 1 2and θ is the arithmetic mean of α &β , then sin2θ + cos2θ is equal to:

Given

$$\mathrm{cos}\alpha +\mathrm{cos}\beta =\frac{3}{2}$$cosα+cosβ=23​

$$\mathrm{sin}\alpha +\mathrm{sin}\beta =\frac{1}{2}$$sinα+sinβ=21​

$\Rightarrow 2cos(\frac{\alpha +\beta }{2})cos(\frac{\alpha -\beta }{2})=\frac{3}{2}$⇒2cos(2α+β​)cos(2α−β​)=23​ ......(i) and

$2\mathrm{sin}⁡(\frac{\alpha +\beta }{2})cos⁡(\frac{\alpha -\beta }{2})=\frac{1}{2}$2sin⁡(2α+β​)cos⁡(2α−β​)=21​ ....(ii)

Dividing (ii) and (i)

$$\mathrm{tan}(\frac{(\alpha +\beta )}{2})$$tan(2(α+β)​)

$$\therefore \mathrm{tan}⁡\theta =\frac{1}{3}$$∴tan⁡θ=31​

$\mathrm{sin}2\theta +\mathrm{cos}⁡2\theta =$sin2θ+cos⁡2θ= $\frac{2\mathrm{tan}⁡\theta }{1+{\mathrm{tan}}^{2}⁡\theta }+ \frac{1-{\mathrm{tan}}^{2}⁡\theta }{1+{\mathrm{tan}}^{2}⁡\theta }=\frac{2(\frac{1}{3})}{1+\frac{1}{9}}+\frac{1-\frac{1}{9}}{1+\frac{1}{9}}$1+tan2⁡θ2tan⁡θ​+1+tan2⁡θ1−tan2⁡θ​=1+91​2(31​)​+1+91​1−91​​

$$= \frac{\frac{2}{3}+\frac{8}{9}}{\frac{10}{9}}=\frac{14}{10}=\frac{7}{5}$$=910​32​+98​​=1014​=57​