Question

In a right angled triangle, acute angles A and B satisfy

$\tan A+\tan B+{\tan }^{2}A+{\tan }^{2}B+{\tan }^{3}A+{\tan }^{3}B=70$

find the angle A and B in radians.

Answer 1

Given- $A+B={90}^0\to \left(1\right)$

Using $A+B={90}^0$ & replace $\tan B$ by $\cot A$, so it becomes

$(\tan A+\cot A)+({\tan }^{2}A+{cot}^{2}A)+({\tan }^{3}A+{cot}^{3}A)=70\to \left(2\right)$

$\because {\tan }^{2}A+{cot}^{2}A={(\tan A+\cot A)}^2-2$

${\tan }^{3}A+{\cot }^{3}A={(\tan A+\cot A)}^3-3(\tan A+\cot A)\to \left(3\right)$

Using $\left(3\right)$ in equation $\left(2\right)$, we get

$x^3+x^2-2x-72=0$,

where $x=\tan A+\cot A$

$\Rightarrow x^3+x^2-2x-72=(x-4)(x^2+5x+18)$

only possible root is $4$

so,

$\tan A+\cot A=4$

solve for $\tan A$-

$\tan A+\frac{1}{\tan A}=4$

$\Rightarrow {\tan }^{2}A-\tan A4+1=0$

$\tan A=\frac{1\pm \sqrt{16-4\left(1\right)\left(1\right)}}{2}$

we get $\tan a=y$

$y=2-\sqrt{3},y=2+\sqrt{3}$

so angle $A=15$ or ${75}^0$ (degrees)

$A=15\left(\frac{\pi }{180}\right)$ or $(\frac{75}{180}\times \pi )$ Radian

$A=\frac{\pi }{12}$ or $\frac{5\pi }{12}$

so $A$ and $B$ are $\frac{5\pi }{12}$ or $\frac{\pi }{12}$