Question

(ctgθ + cscθ)²≡ 1+ cos÷ 1- cosθ

Answer 1

$(cotg(x)+cosec(x))^2=\left(\frac{cos(x)}{sen(x)}+\frac1{sen(x)}\right)=$
$=\left(\frac{1+cos(x)}{sen(x)}\right)^2=$
$=\frac{(1+cos(x))^2}{sen^2(x)}=$
$=\frac{(1+cos(x))^2}{1-cos^2(x)}=$
$=\frac{(1+cos(x))^2}{(1+cos(x))(1-cos(x))}$
$=\frac{1+cos(x)}{1-cos(x)}$