# I curious practical solution.(Step by step) ( 1 + tan &#x2061;<!-- ⁡ -->

I curious practical solution.(Step by step)
$\left(1+\mathrm{tan}{5}^{\circ }\right)\left(1+\mathrm{tan}{10}^{\circ }\right)\left(1+\mathrm{tan}{15}^{\circ }\right)\cdots \left(1+\mathrm{tan}{40}^{\circ }\right)$
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If $A+B=\frac{\pi }{4}\phantom{\rule{thickmathspace}{0ex}},$ Then $\left(1+\mathrm{tan}A\right)\left(1+\mathrm{tan}B\right)=2$
So $\left(1+\mathrm{tan}{5}^{0}\right)\left(1+\mathrm{tan}{40}^{0}\right)=2$
and $\left(1+\mathrm{tan}{15}^{0}\right)\left(1+\mathrm{tan}{30}^{0}\right)=2$
So
$\left(1+\mathrm{tan}{5}^{0}\right)\left(1+\mathrm{tan}{40}^{0}\right)\left(1+\mathrm{tan}{10}^{0}\right)\left(1+\mathrm{tan}{35}^{0}\right)\left(1+\mathrm{tan}{15}^{0}\right)\left(1+\mathrm{tan}{30}^{0}\right)\left(1+\mathrm{tan}{20}^{0}\right)\left(1+\mathrm{tan}{35}^{0}\right)={2}^{4}=16$
###### Not exactly what you’re looking for?
Hayley Sanders
For example, observe that
$1=\mathrm{tan}{45}^{\circ }=\mathrm{tan}\left({5}^{\circ }+{40}^{\circ }\right)=\frac{\mathrm{tan}{5}^{\circ }+\mathrm{tan}{40}^{\circ }}{1-\mathrm{tan}{5}^{\circ }\mathrm{tan}{40}^{\circ }}.$
Then we have $\mathrm{tan}{5}^{\circ }+\mathrm{tan}{40}^{\circ }=1-\mathrm{tan}{5}^{\circ }\mathrm{tan}{40}^{\circ }$, and thus
$\begin{array}{rl}\left(1+\mathrm{tan}{5}^{\circ }\right)\left(1+\mathrm{tan}{40}^{\circ }\right)& =1+\mathrm{tan}{5}^{\circ }\mathrm{tan}{40}^{\circ }+\mathrm{tan}{5}^{\circ }+\mathrm{tan}{40}^{\circ }\\ & =1+\mathrm{tan}{5}^{\circ }\mathrm{tan}{40}^{\circ }+1-\mathrm{tan}{5}^{\circ }\mathrm{tan}{40}^{\circ }\\ & =2.\end{array}$
Hence we conclude that
$\left(1+\mathrm{tan}{5}^{\circ }\right)\left(1+\mathrm{tan}{10}^{\circ }\right)\cdots \left(1+\mathrm{tan}{40}^{\circ }\right)={2}^{4}=16.$