A geometry problem by Nazmus sakib

Geometry Level 2

$\large \tan^{-1} \frac {1}{2} + \tan^{-1} \frac{ 1}{3} = ?$

\dfrac{\pi}{2}

\dfrac{\pi}{8}

\dfrac{\pi}{10}

\dfrac{\pi}{6}

\dfrac{\pi}{4}

1 solution

Chew-Seong Cheong
Sep 9, 2017

$\begin{aligned} \theta & = \tan^{-1} \frac 12 + \tan^{-1} \frac 13 \\ \tan \theta & = \tan \left(\tan^{-1} \frac 12 + \tan^{-1} \frac 13\right) \\ & = \frac {\frac 12 + \frac 13}{1-\frac 12 \times \frac 13} = 1 \\ \implies \theta & = {\color{#3D99F6}n}\pi + \boxed{\dfrac \pi 4} & \small \color{#3D99F6} \text{where }n \text{ is an integer.} \end{aligned}$

@Sakib Nazmus , you can just enter \tan for $\tan$ . Similarly, \sin $\sin$ , \cos $\cos$ , \sinh $\sinh$ , \In $\ln$ , \log $\log$ , \gcd $\gcd$ , \int $\int$ , \sum $\sum$ . When you are using \ [ \ ], don't need to use \dfrac just \frac will do also no need \displaystyle. When the operand is only one character. braces { } are unnecessary, for example, \frac 12 $\frac 12$ , \sin^2 $\sin^2$ , a_1 $a_1$ .

Chew-Seong Cheong - 3 years, 9 months ago

Sir, thanks for the information.

Nazmus sakib - 3 years, 9 months ago

Put your mouse cursor on top of the formulas and you will see the LaTex codes. You can also click the pull-down menu ( $\cdots$ ) at the right bottom corner of the problem and select Toggle LaTex to see the codes.

Chew-Seong Cheong - 3 years, 9 months ago

A geometry problem by Nazmus sakib

1 solution

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