If $11 \sin^{2} \theta - \cos^{2} \theta + 4 \sin \theta - 4 = 0, 0^\circ < \theta < 90^\circ$, then what is the value of $\frac{\cos 2\theta + \cot 2 \theta}{\sec 2 \theta - \tan 2 \theta}$
Answer & explanation
Correct answer: option 1
11 sin²θ - cos²θ + 4 sinθ - 4 = 0
{ sin²θ + cos²θ = 1 }
11 sin²θ - (1 - sin²θ) + 4 sinθ - 4 = 0
11 sin²θ - 1 + sin²θ + 4 sinθ - 4 = 0
12 sin²θ + 4 sinθ - 5 = 0
12 sin²θ + 10 sinθ - 6sinθ - 5 = 0
2sinθ ( 6sinθ +5 ) - 1 ( 6sinθ +5 ) = 0
( 2sinθ - 1 ). ( 6sinθ +5 ) = 0
Either ( 2sinθ - 1 ) = 0 Or ( 6sinθ +5 ) = 0
( 6sinθ +5 ) = 0 is not possible.
So, 2sinθ - 1 = 0
sinθ = \(\frac{1}{2}\)
{ sin30º = \(\frac{1}{2}\) }
So, θ = 30º
Now,
\(\frac{sin2θ + cot2θ }{sec2θ - tan2θ }\)
= \(\frac{sin60º + cot60º}{sec60º - tan60º }\)
= \(\frac{1/2 + √3}{2 -√3 }\)
= \(\frac{2 + √3}{2√3(2 -√3) }\)
= \(\frac{2 + √3}{2√3(2 -√3) }\) × \(\frac{2 + √3}{2 +√3}\)
= \(\frac{12 + 7√3}{6 }\)