JEE Main 2021 · 25 July · Evening Shift · Mathematics · Trigonometric Ratios & Identities
B. 2+3+2+6\sqrt{2} + \sqrt{3} + 2 + \sqrt{6}2+3+2+6
We know that cotθ=1+cos2θsin2θcot\theta = \dfrac{1 + \cos 2\theta}{\sin 2\theta}cotθ=sin2θ1+cos2θ
Since, θ=π24\theta = \dfrac{\pi}{24}θ=24π
Therefore, cotθ=1+(3+122)(3−122)cot\theta = \dfrac{1 + \left(\dfrac{\sqrt{3} + 1}{2\sqrt{2}}\right)}{\left(\dfrac{\sqrt{3} - 1}{2\sqrt{2}}\right)}cotθ=(223−1)1+(223+1)
⇒cot(π24)=1+(3+122)(3−122)⇒ \cot \left(\dfrac{\pi}{24}\right) = \dfrac{1 + \left(\dfrac{\sqrt{3} + 1}{2\sqrt{2}}\right)}{\left(\dfrac{\sqrt{3} - 1}{2\sqrt{2}}\right)}⇒cot(24π)=(223−1)1+(223+1)
=(22+3+1)(3−1)×(3+1)(3+1)= \dfrac{\left(2\sqrt{2} + \sqrt{3} + 1\right)}{\left(\sqrt{3} - 1\right)} \times \dfrac{\left(\sqrt{3} + 1\right)}{\left(\sqrt{3} + 1\right)}=(3−1)(22+3+1)×(3+1)(3+1)
=26+22+3+3+3+12= \dfrac{2\sqrt{6} + 2\sqrt{2} + 3 + \sqrt{3} + \sqrt{3} + 1}{2}=226+22+3+3+3+1
=6+2+3+2.= \sqrt{6} + \sqrt{2} + \sqrt{3} + 2 .=6+2+3+2.