glamrockqueen7

2021-08-15

If $\theta$ is an acute angle and $\mathrm{tan}\theta +\mathrm{cot}\theta =2$. Find the value of ${\mathrm{tan}}^{7}\theta +{\mathrm{cot}}^{7}\theta$

coffentw

Given:
$\mathrm{tan}\theta +\mathrm{cot}\theta =2$ and Q is acute angle
${\mathrm{tan}}^{7}\theta +{\mathrm{cot}}^{7}\theta$
$\mathrm{tan}\theta +\mathrm{cot}\theta =2$
$\mathrm{tan}\theta +\frac{1}{\mathrm{tan}\theta }=2$
$\frac{{\mathrm{tan}}^{2}\theta +1}{\mathrm{tan}\theta }=2⇒{\mathrm{tan}}^{2}\theta +1=2\mathrm{tan}\theta$
${\mathrm{tan}}^{2}\theta -2\mathrm{tan}\theta +1=0$
$\mathrm{tan}\theta \left(\mathrm{tan}\theta -1\right)-1\left(\mathrm{tan}\theta -1\right)=0$
$\left(\mathrm{tan}\theta -1\right)-1\left(\mathrm{tan}\theta -1\right)=0$
$\mathrm{tan}\theta -1=0⇒\mathrm{tan}\theta =1$
$\mathrm{tan}\theta =1$
$\theta ={95}^{\circ }$
So,
$\mathrm{cot}\theta =\frac{1}{\mathrm{tan}\theta }=\frac{1}{1}=1$
Now, ${\mathrm{tan}}^{7}\theta +{\mathrm{cot}}^{7}\theta$
$={\left(\mathrm{tan}\theta \right)}^{7}+{\left(\mathrm{cot}\theta \right)}^{7}$
$={1}^{7}+{1}^{7}$
$=1+1$
${\mathrm{tan}}^{7}+{\mathrm{cot}}^{7}=2$

Jeffrey Jordon