If $ \cot a + \frac 1 {\cot a} = 1 $, then what is $ \cot^2 a + \frac 1{\cot^2 a}$?
the answer is given as $-1$ in my book, but how do you arrive at this conclusion?
If $ \cot a + \frac 1 {\cot a} = 1 $, then what is $ \cot^2 a + \frac 1{\cot^2 a}$?
the answer is given as $-1$ in my book, but how do you arrive at this conclusion?
cota+ (1/cota)=1
Therefore, Squaring on both sides we get:
cot^2a + (1/cot^2a)+ 2 = 1 Hence, cot^2a + (1/cot^2a) = -1
Hint: $ x^2+\frac1{x^2}=\left(x+\frac1x\right)^2-2. $
Hint What do you get if you square $\cot(a)+\frac{1}{\cot(a)}$?
Taking $x=\cot a$, $x+\frac{1}{x}=1\implies x^2+\frac{1}{x^2}=(x+\frac{1}{x})^2-2x\frac{1}{x}=1-2=-1$
Alternatively, $x+\frac{1}{x}=1\implies x^2-x+1=0$
$x^2-x+1=0\implies x^3+1=0$
So,
$x^{3m}+(\frac{1}{x})^{3m}=(x^3)^m+\frac{1}{(x^3)^m}=2(-1)^m$
$x^{3m+1}+(\frac{1}{x})^{3m+1}=(x^3)^m\cdot x+\frac{1}{(x^3)^m\cdot x}=(-1)^m(x+\frac{1}{x})=(-1)^m$
$x^{3m+2}+(\frac{1}{x})^{3m+2}=(x^3)^m\cdot x^2+\frac{1}{(x^3)^m\cdot x^2}=(-1)^m(x^2+\frac{1}{x^2})=(-1)^m(-\frac{1}{x}-x)$ as $x^3=-1\implies x^2=-\frac{1}{x}$ and $\frac{1}{x^2}=x$
So,$x^{3m+2}+(\frac{1}{x})^{3m+2}=(-1)^{m+1}\cdot (x+\frac{1}{x})=(-1)^{m+1}$
If we put $m=0$, $x^2+\frac{1}{x^2}=(-1)^1=-1$