# If tan &#x2061;<!-- ⁡ --> ( &#x03B1;<!-- α --> ) + cot &#x2061;<!-- ⁡ -->

If $\mathrm{tan}\left(\alpha \right)+\mathrm{cot}\left(\alpha \right)=p,$, denote ${\mathrm{tan}}^{3}\left(\alpha \right)+{\mathrm{cot}}^{3}\left(\alpha \right)$ in terms of p
You can still ask an expert for help

## Want to know more about Trigonometry?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Nicolas Calhoun
Let $u=\mathrm{tan}\left(\alpha \right)$
$\begin{array}{rl}\text{Solu}& \text{tion #1:}\\ & u+\frac{1}{u}=p\\ \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}& {\left(u+\frac{1}{u}\right)}^{2}={p}^{2}\\ \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}& {u}^{2}+\frac{1}{{u}^{2}}={p}^{2}-2\\ \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}& {u}^{2}-u\left(\frac{1}{u}\right)+\frac{1}{{u}^{2}}={p}^{2}-3\\ \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}& \left(u+\frac{1}{u}\right)\left({u}^{2}-u\left(\frac{1}{u}\right)+\frac{1}{{u}^{2}}\right)=p\left({p}^{2}-3\right)\\ \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}& {u}^{3}+{\left(\frac{1}{u}\right)}^{3}=p\left({p}^{2}-3\right)\\ \text{Solu}& \text{tion #2:}\\ & u+\frac{1}{u}=p\\ \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}& {\left(u+\frac{1}{u}\right)}^{3}={p}^{3}\\ \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}& {u}^{3}+3{u}^{2}\left(\frac{1}{u}\right)+3u{\left(\frac{1}{u}\right)}^{2}+{\left(\frac{1}{u}\right)}^{3}={p}^{3}\\ \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}& {u}^{3}+3\left(u+\frac{1}{u}\right)+{\left(\frac{1}{u}\right)}^{3}={p}^{3}\\ \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}& {u}^{3}+{\left(\frac{1}{u}\right)}^{3}={p}^{3}-3p\end{array}$