If $8sin^2θ$+ 2 cosθ = 5, 0° < θ < 90°, the the value of $tan^2θ + sec^2θ - sin^2θ$ will be :
Answer & explanation
Correct answer: option 1
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
= (\(\frac{7}{9}\)) + ( \(\frac{16}{9}\) ) - ( \(\frac{7}{16}\) )
= ( \(\frac{305}{144}\) )