JEE Main 20265 Apr, Morning shiftMathematicsEasyTrigonometry118× since 2002Let tanA,tanB\tan A, \tan BtanA,tanB, where A,B∈(−π2,π2)A, B \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)A,B∈(−2π,2π), be the roots of the quadratic equation x2−2x−5=0x^2 - 2x - 5 = 0x2−2x−5=0. Then 20sin2(A+B2)20\sin^2\left(\frac{A+B}{2}\right)20sin2(2A+B) is equal to:AOption A. 10+1010+\sqrt{10}10+10BOption B. 10−21010-2\sqrt{10}10−210COption C. 10−31010-3\sqrt{10}10−310DOption D. 10−1010-\sqrt{10}10−10CheckSkip · show solution
