arctan 12\frac{1}{2}21 + arctan 13\frac{1}{3}31=x
tanx + sin2sin^{2}sin2x=?
asked by guest on Dec 08, 2024 at 2:25 am
MathBot Answer:
sin2(x)+tan(x)=sin2(arctan(13)+arctan(12))+tan(arctan(13)+arctan(12))\sin^{2}{\left(x \right)} + \tan{\left(x \right)} = \sin^{2}{\left(\arctan{\left(\frac{1}{3} \right)} + \arctan{\left(\frac{1}{2} \right)} \right)} + \tan{\left(\arctan{\left(\frac{1}{3} \right)} + \arctan{\left(\frac{1}{2} \right)} \right)}sin2(x)+tan(x)=sin2(arctan(31)+arctan(21))+tan(arctan(31)+arctan(21))
atan(13)+atan(12)=x−x=−atan(12)−atan(13)x=atan(13)+atan(12)\begin{aligned}\operatorname{atan}{\left(\frac{1}{3} \right)} + \operatorname{atan}{\left(\frac{1}{2} \right)} &= x\\- x &= - \operatorname{atan}{\left(\frac{1}{2} \right)} - \operatorname{atan}{\left(\frac{1}{3} \right)}\\x &= \operatorname{atan}{\left(\frac{1}{3} \right)} + \operatorname{atan}{\left(\frac{1}{2} \right)}\end{aligned}atan(31)+atan(21)−xx=x=−atan(21)−atan(31)=atan(31)+atan(21)
By substituting this value into the expression, we find that: sin2(x)+tan(x)=sin2(arctan(13)+arctan(12))+tan(arctan(13)+arctan(12))=sin2(arctan(13)+arctan(12))+tan(arctan(13)+arctan(12))\begin{aligned}\sin^{2}{\left(x \right)} + \tan{\left(x \right)} &= \sin^{2}{\left(\arctan{\left(\frac{1}{3} \right)} + \arctan{\left(\frac{1}{2} \right)} \right)} + \tan{\left(\arctan{\left(\frac{1}{3} \right)} + \arctan{\left(\frac{1}{2} \right)} \right)}\\&= \sin^{2}{\left(\arctan{\left(\frac{1}{3} \right)} + \arctan{\left(\frac{1}{2} \right)} \right)} + \tan{\left(\arctan{\left(\frac{1}{3} \right)} + \arctan{\left(\frac{1}{2} \right)} \right)}\end{aligned}sin2(x)+tan(x)=sin2(arctan(31)+arctan(21))+tan(arctan(31)+arctan(21))=sin2(arctan(31)+arctan(21))+tan(arctan(31)+arctan(21))