For θ : 0° < θ < 90°
$3 sec θ + 4 cos θ = 4 \sqrt{3}$, find the value of ( 1 - sin θ + cos θ)
Answer & explanation
Correct answer: option 2
3secθ + 4 cosθ = 4√3
3( \(\frac{1}{cosθ}\)) + 4 cosθ = 4√3
3 + 4 cos²θ = 4√3 cosθ
4 cos²θ - 4√3 cosθ + 3 = 0
4 cos²θ - 2√3 cosθ - 2√3 cosθ + 3 = 0
2 cosθ( 2cosθ - √3 ) - √3( 2cosθ - √3 ) = 0
( 2cosθ - √3 ). ( 2cosθ - √3 ) = 0
2cosθ - √3 = 0
cosθ = \(\frac{√3}{2}\)
{ cos 30º = \(\frac{√3}{2}\) }
Now,
1 - sinθ + cosθ
= 1 - sin30º + cos30º
= 1 - \(\frac{1}{2}\) + \(\frac{√3}{2}\)
= \(\frac{1 + √3}{2}\)