If $12 \cos^2 \theta - 2 \sin^2 \theta + 3\cos \theta = 3, 0^\circ < \theta < 90^\circ$, then what is the value of $\frac{cosec~\theta + \sec \theta}{\tan \theta + \cot \theta}$?
Answer & explanation
Correct answer: option 3
12 cos²θ - 2sin²θ + 3cosθ = 3
12 cos²θ - 2 ( 1 - cos²θ ) + 3cosθ = 3
14 cos²θ + 3cosθ - 5 = 0
14 cos²θ + 10cosθ -7cosθ - 5 = 0
2cosθ ( 7cosθ + 5 ) - 1 ( 7cosθ + 5 ) = 0
( 2cosθ - 1 ) . ( 7cosθ + 5 ) = 0
Either ( 2cosθ - 1 ) = 0 Or ( 7cosθ + 5 ) = 0
( 7cosθ + 5 ) = 0
cosθ = - \(\frac{5}{7}\) ( not possible )
So, ( 2cosθ - 1 ) = 0
cosθ = \(\frac{1}{2}\)
{ we know, cos60º = \(\frac{1}{2}\) }
So , θ = 60º
Now,
\(\frac{cosecθ+ secθ }{tanθ + cotθ}\)
= \(\frac{cosec60º+ sec60º }{tan60º + cot60º}\)
= \(\frac{2/√3 +2 }{√3 + 1/√3}\)
= \(\frac{2√3 +2 }{4}\)
= \(\frac{√3 +1 }{2}\)