If $3 + \cos^2 \theta = 3(\cot^2 \theta + \sin^2 \theta), 0^\circ < \theta < 90^\circ$, then what is the value of $(\cos \theta + 2 \sin \theta)$?
Answer & explanation
Correct answer: option 3
3 + cos²θ = 3 ( cot²θ + sin²θ )
Put θ = 30º
3 + cos²θ = 3 ( cot²θ + ( 1 - cos²θ ) )
3 + cos²θ = 3cot²θ + 3( 1 - cos²θ )
4cos²θ = 3cot²θ
Sinθ = \(\frac{√3}{2}\)
{ We know, Sin60º = \(\frac{√3}{2}\) }
So, θ = 60º
Now,
cosθ + 2sinθ
= cos60º + 2sin60º
= \(\frac{1}{2}\) + 2 × \(\frac{√3}{2}\)
= \(\frac{1}{2}\) + √3
= \(\frac{1 +2√3 }{2}\)