JEE Main 2020 — Mathematics Trigonometry

Given,

$\frac{\sqrt{2}\mathrm{sin}\alpha }{\sqrt{1+\mathrm{cos}2\alpha }}=\frac{1}{7}$1+cos2α​2​sinα​=71​ and $\sqrt{\frac{1-\mathrm{cos}2\beta }{2}}=\frac{1}{\sqrt{10}},\alpha , \beta \in (0,\frac{\pi }{2})$21−cos2β​​=10​1​,α,β∈(0,2π​)

Now using $\mathrm{cos}2\theta =2{\mathrm{cos}}^{2}\theta -1=1-2{\mathrm{sin}}^{2}\theta$cos2θ=2cos2θ−1=1−2sin2θ, we can write

$\frac{\sqrt{2}\mathrm{sin}⁡\alpha }{\sqrt{2}\mathrm{cos}⁡\alpha }=\frac{1}{7}$2​cos⁡α2​sin⁡α​=71​ and $\frac{\sqrt{2}\mathrm{sin}⁡\beta }{\sqrt{2}}=\frac{1}{\sqrt{10}}$2​2​sin⁡β​=10​1​

tan⁡\alpha =17 and $\mathrm{sin}⁡\beta =\frac{1}{\sqrt{10}}$sin⁡β=10​1​ or $\mathrm{tan}⁡\beta =\frac{1}{3}$tan⁡β=31​

$$tan2\beta =\frac{2\mathrm{tan}⁡\beta }{1-{\mathrm{tan}}^{2}⁡\beta }=2.\frac{(\frac{1}{3})}{[1-(\frac{1}{9})]}=\frac{3}{4}$$tan2β=1−tan2⁡β2tan⁡β​=2.[1−(91​)](31​)​=43​

$$\mathrm{tan}⁡(\alpha +2\beta )=\frac{tan\alpha +tan2\beta }{1-tan\alpha \mathrm{tan}⁡2\beta }=\frac{\frac{1}{7}+\frac{3}{4}}{1-\frac{1}{7}\cdot \frac{3}{4}}=\frac{\frac{4+21}{28}}{\frac{25}{28}}=1$$tan⁡(α+2β)=1−tanαtan⁡2βtanα+tan2β​=1−71​⋅43​71​+43​​=2825​284+21​​=1