If $8sin^2θ$+ 2 cosθ = 5, 0° < θ < 90°, the the value of $tan^2θ + sec^2θ - sin^2θ$ will be :

$\frac{305}{144}$

8 sin²θ + 2 cosθ = 5

{ sin²θ + cos²θ = 1 }

8(1 - cos²θ ) + 2 cosθ = 5

8cos²θ - 2cosθ + 3 = 0

8cos²θ - 6cosθ + 4cosθ + 3 = 0

2cosθ ( 4cosθ - 3 ) + 1 ( 4cosθ - 3 )= 0

( 2cosθ - 1 ) . ( 4cosθ - 3 )= 0

Either 2cosθ + 3 = 0 or 4cosθ - 3  = 0

cosθ = - \(\frac{1}{2}\)    is not possible.

So, 4cosθ - 3  = 0 

cosθ = \(\frac{3}{4}\)

{ cosθ = \(\frac{B}{H}\) }

By using pythagoras theorem ,

P² + B² = H²

P² + 3² = 4²

P = √7

Now,

tan²θ + sec²θ - sin²θ

= (\(\frac{√7}{3}\))² +  ( \(\frac{4}{3}\) )² -  ( \(\frac{√7}{4}\) )²

= (\(\frac{7}{9}\)) +  ( \(\frac{16}{9}\) ) -  ( \(\frac{7}{16}\) )

= ( \(\frac{305}{144}\) )