Verified Solution • 2025–26 Session moderate

QJEEM21Macbff: JEE Main 2021 Mathematics - Class 11 JEEM21Macbff Solution: Let tan , tan and tan ; , , , n N be the slopes of three line…

Let tan α\alpha , tan β\beta and tan γ\gamma ; α\alpha , β\beta , γ\gamma ≠\ne (2n−1)π2{{(2n - 1)\pi } \over 2} , n ∈\in N be the slopes of three line segments OA, OB and OC, respectively, where O is origin. If circumcentre of Δ\Delta ABC coincides with origin and its orthocentre lies on y-axis, then the value of (cos⁡3α+cos⁡3β+cos⁡3γcos⁡αcos⁡βcos⁡γ)2{\left( {{{\cos 3\alpha + \cos 3\beta + \cos 3\gamma } \over {\cos \alpha \cos \beta \cos \gamma }}} \right)^2} is equal to ____________.

Step-by-Step Verified Solution

Since orthocentre and circumcentre both lies on y-axis.

⇒\Rightarrow Centroid also lies on y-axis.

⇒\Rightarrow ∑cos⁡α=0\sum {\cos \alpha = 0}

cos α\alpha + cos β\beta + cos γ\gamma = 0

⇒\Rightarrow cos3 α\alpha + cos3 β\beta + cos3 γ\gamma = 3cos α\alpha cos β\beta cos γ\gamma

∴\therefore cos⁡3α+cos⁡3β+cos⁡3γcos⁡αcos⁡βcos⁡γ{{\cos 3\alpha + \cos 3\beta + \cos 3\gamma } \over {\cos \alpha \cos \beta \cos \gamma }}

=4(cos⁡3α+cos⁡3β+cos⁡3γ)−3(cos⁡α+cos⁡β+cos⁡γ)cos⁡αcos⁡βcos⁡γ=12= {{4({{\cos }^3}\alpha + {{\cos }^3}\beta + {{\cos }^3}\gamma ) - 3(\cos \alpha + \cos \beta + \cos \gamma )} \over {\cos \alpha \cos \beta \cos \gamma }} = 12

then, (cos⁡3α+cos⁡3β+cos⁡3γcos⁡αcos⁡βcos⁡γ)2=144{\left( {{{\cos 3\alpha + \cos 3\beta + \cos 3\gamma } \over {\cos \alpha \cos \beta \cos \gamma }}} \right)^2} = 144

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