arctan 12\frac{1}{2} + arctan 13\frac{1}{3}=x

tanx + sin2sin^{2}x=?

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)}


atan(13)+atan(12)=xx=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}

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}