If $$ (1 + \tan^2 \theta) + (1 + ( \tan^2 \theta)^{-1}) = k,  then  \sqrt k = ?$$

$$ (1 + \tan^2 \theta) + (1 + ( \tan^2 \theta)^{-1})$$ = k

= where θ is replaced by  x

use this identity to solve this problem tanx = $$\frac{sinx}{cosx}$$

from the question

= (1+$$\frac{sin^2x}{cos^2x}$$) + (1+$$\frac{1}{tan^2x}$$) = k

=  ($$\frac{cos^2x+sin^2x}{cos^2x}$$)+(1+$$\frac{1}{sin^2x÷cos^2x}$$) = k

= use this trigonometric identity $$\cos^2 \theta + \sin^2 \theta$$ = 1

= ($$\frac{1}{cos^2x}$$) + (1+$$\frac{cos^2x}{sin^2x}$$) = k

= ($$\frac{1}{cos^2x}$$) + ($$\frac{sin^2x+cos^2x}{sin^2x}$$) = k

= ($$\frac{1}{cos^2x}$$) + ($$\frac{1}{sin^2x}$$) take the order of this then

= on further simplification it becomes

= ($$\frac{cos^2x+sin^2x}{cos^2x × sin^2x}$$) = k

= ($$\frac{1}{cos^2x × sin^2x}$$) = k

we need to find √k hence after squaring we get = $$\frac{1}{cosx × sinx}$$

where  $$\frac{1}{cosx}$$ = secx  AND   $$\frac{1}{sinx}$$   =  cosecx from the trigonometric inverse functions

 Hence the solution is  cosecx secx