Question

find the calculation

Answer 1

refer the attachment

Answer 2

sec x can also be written as 1/tanx
hence
$1 \div \tan(x) + \tan(x) = 3$
$= 1 + \tan(x) {}^{2} \div \tan(x ) = 3$
we know that 1/tanx =sec x
and 1+tan square x = sec square x
hence
$\sec(x) {}^{2} \div \tan(x ) = 3$
$= \sec(x) {}^{3} = 3$
hence
$\sec(x) = \sqrt[3]{3}$