$(\sec \theta+\tan \theta)^2+\frac{1+{cosec} \theta}{1-{cosec} \theta}, 0^{\circ}<\theta<90^{\circ} \text { is: }$

0

( secθ + tanθ )² + \(\frac{1 + cosecθ}{1 - cosecθ}\)

Let us assume that ,

θ = 45º

= ( sec45º + tan45º )² + \(\frac{1 + cosec45º}{1 - cosec45º}\)

= ( √2 + 1 )² + \(\frac{1 + √2}{1 - √2}\)

= ( √2 + 1 )² + \(\frac{1 + √2}{1 - √2}\) × \(\frac{1 + √2}{1 + √2}\)

= ( √2 + 1 )² + \(\frac{( √2 + 1 )²}{1 - (√2)²}\)

= ( √2 + 1 )² + \(\frac{( √2 + 1 )²}{-1}\)

= ( √2 + 1 )² - ( √2 + 1 )²

= 0