If ${\tan }^3\theta +{\cot }^3\theta =52,$  then ${\tan }^2\theta +{\cot }^2\theta =\lambda$  where $\frac{\lambda }{2}$  equals to........

   ${\tan }^3\theta +{\cot }^3\theta =52$
$\Rightarrow (x+\frac{1}{x})(x^2-1+\frac{1}{x^2})=52$  or  ${(x+\frac{1}{x})}^3-3(x+\frac{1}{x})-52=0$
$\Rightarrow x+\frac{1}{x}=4\Rightarrow x^2+\frac{1}{x^2}={(x+\frac{1}{x})}^2-2=14,$   $x$  being  $\tan \theta$