arctan $\frac{1}{2}$ + arctan $\frac{1}{3}$=x

tanx + $sin^{2}$x=?

asked by guest
on Dec 08, 2024 at 2:25 am



You asked:

Given \(\arctan\left( \frac{1}{2} \right) + \arctan\left( \frac{1}{3} \right) = x\) evaluate the expression \(\tan\left( x \right) + {\sin\left( x \right)}^{2}\).

MathBot Answer:

\[\sin^{2}{\left(x \right)} + \tan{\left(x \right)} = \sin^{2}{\left(\arctan{\left(\frac{1}{3} \right)} + \arctan{\left(\frac{1}{2} \right)} \right)} + \tan{\left(\arctan{\left(\frac{1}{3} \right)} + \arctan{\left(\frac{1}{2} \right)} \right)}\]

\[\begin{aligned}\operatorname{atan}{\left(\frac{1}{3} \right)} + \operatorname{atan}{\left(\frac{1}{2} \right)} &= x\\- x &= - \operatorname{atan}{\left(\frac{1}{2} \right)} - \operatorname{atan}{\left(\frac{1}{3} \right)}\\x &= \operatorname{atan}{\left(\frac{1}{3} \right)} + \operatorname{atan}{\left(\frac{1}{2} \right)}\end{aligned}\]

By substituting this value into the expression, we find that: $$\begin{aligned}\sin^{2}{\left(x \right)} + \tan{\left(x \right)} &= \sin^{2}{\left(\arctan{\left(\frac{1}{3} \right)} + \arctan{\left(\frac{1}{2} \right)} \right)} + \tan{\left(\arctan{\left(\frac{1}{3} \right)} + \arctan{\left(\frac{1}{2} \right)} \right)}\\&= \sin^{2}{\left(\arctan{\left(\frac{1}{3} \right)} + \arctan{\left(\frac{1}{2} \right)} \right)} + \tan{\left(\arctan{\left(\frac{1}{3} \right)} + \arctan{\left(\frac{1}{2} \right)} \right)}\end{aligned}$$

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