If $(cos^2 \theta -1) (2sec^2 \theta) + sec^2 \theta + 2tan^2\theta = 2, 0° < \theta < 90°$, then the value of $\frac{(sec\theta+sin\theta)}{(cosec\theta - cos \theta)}$ will be :

3

( cos²θ  - 1 ) (2sec²θ ) + sec²θ  + 2 tan²θ  = 2 

{using formula, sin²θ  + cos²θ  = 1 }

( -sin²θ   ) (2sec²θ ) + sec²θ  + 2 tan²θ  = 2

( -2tan²θ ) + sec²θ  + 2 tan²θ  = 2

{using formula, sec²θ  -  tan²θ  = 1 }

  tan²θ  = 1

{we know, tan45º = 1}

S0 ,  θ = 45º

Now,

\(\frac{secθ+ sinθ}{cosecθ - cosθ}\)

= \(\frac{sec45º+ sin45º}{cosec45º - cos45º}\)

= \(\frac{√2+ 1/√2}{√2 - 1/√2}\)

= \(\frac{3}{1}\)

= 3