Question:
If \(4\cos A + 2\sec A = 4\sqrt{2}\), where \(0^\circ < A < 90^\circ\), then find the value of angle A.
यदि \(4\cos A + 2\sec A = 4\sqrt{2}\) है, जहाँ \(0^\circ < A < 90^\circ\) है, तो कोण A का मान ज्ञात कीजिए।
Options:
- 30°
- 45°
- 60°
- 15°
- 30°
- 45°
- 60°
- 15°
Correct Answer: b)
Explanation:
Substitute \(\sec A = \dfrac{1}{\cos A}\):
\(4\cos A + \dfrac{2}{\cos A} = 4\sqrt{2} \implies 4\cos^2 A – 4\sqrt{2}\cos A + 2 = 0\).
Dividing by 2: \(2\cos^2 A – 2\sqrt{2}\cos A + 1 = 0 \implies (\sqrt{2}\cos A – 1)^2 = 0\).
\(\cos A = \dfrac{1}{\sqrt{2}} \implies A = 45^\circ\).
\(4\cos^2 A – 4\sqrt{2}\cos A + 2 = 0 \implies (\sqrt{2}\cos A – 1)^2 = 0 \implies \cos A = \dfrac{1}{\sqrt{2}}\)।
अतः \(A = 45^\circ\)।