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Tan 2A=cot(A-15) Or, Tan 2A =Tan [90-(A-15)] Or,Tan2A=Tan(90+15-A) Or,Tan2A=Tan(105-A) This implies, 2A= If tanA+tan2A+√3tanAtan2A=√3 then general solution of A2= · Solve tanθ+tan2θ+√3tanθtan2θ=√3. Anπ+π3. Bnπ2+π6 · If 2tan(π/3)cos(2πx)=√3, the general solution of ... Prove the following trigonometric identity: cot (45° -A) = tan 2A + sec 2A. Solution: LHS = cot (45° -A) = \dfrac{cot45°.cotA + 1}{cotA May 21, 2020 ... If sinA=(3/5) and A is in first quadrant, then the values of sin2A, cos2A and tan2A are (A) 24/25, 7/25, 24/7 (B) 1/25, 7/25, 1/7 (C) 24/25, ... l +tan2A. (13) where £/

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