Question:
Let \(\sin\beta\) be the GM of \(\sin\alpha\) and \(\cos\alpha\); \(\tan\gamma\) be the AM of \(\sin\alpha\) and \(\cos\alpha\). What is the value of \(\sec 2\gamma\)?
मान लीजिए \(\sin\beta\), \(\sin\alpha\) और \(\cos\alpha\) का गुणोत्तर माध्य (GM) है; और \(\tan\gamma\), \(\sin\alpha\) और \(\cos\alpha\) का समांतर माध्य (AM) है। \(\sec 2\gamma\) का मान क्या है?
Options:
- \(\dfrac{3 – 2\sin 2\alpha}{4 + \sin 2\alpha}\)
- \(\dfrac{3 – \sin 2\alpha}{4 + 3\sin 2\alpha}\)
- \(\dfrac{5 + \sin 2\alpha}{3 – \sin 2\alpha}\)
- \(\dfrac{3 – \sin 2\alpha}{5 + 2\sin 2\alpha}\)
- \(\dfrac{3 – 2\sin 2\alpha}{4 + \sin 2\alpha}\)
- \(\dfrac{3 – \sin 2\alpha}{4 + 3\sin 2\alpha}\)
- \(\dfrac{5 + \sin 2\alpha}{3 – \sin 2\alpha}\)
- \(\dfrac{3 – \sin 2\alpha}{5 + 2\sin 2\alpha}\)
Correct Answer: c)
Explanation:
Here \(\sin\beta = \sqrt{\sin\alpha\cos\alpha}\) and \(\tan\gamma = \dfrac{\sin\alpha+\cos\alpha}{2}\).
\(\sec 2\gamma = \dfrac{1+\tan^2\gamma}{1-\tan^2\gamma} = \dfrac{1 + \frac{(\sin\alpha+\cos\alpha)^2}{4}}{1 – \frac{(\sin\alpha+\cos\alpha)^2}{4}} = \dfrac{4 + (1 + 2\sin\alpha\cos\alpha)}{4 – (1 + 2\sin\alpha\cos\alpha)} = \dfrac{5 + \sin 2\alpha}{3 – \sin 2\alpha}\).
यहाँ \(\sin\beta = \sqrt{\sin\alpha\cos\alpha}\) और \(\tan\gamma = \dfrac{\sin\alpha+\cos\alpha}{2}\) हैं।
\(\sec 2\gamma = \dfrac{1+\tan^2\gamma}{1-\tan^2\gamma} = \dfrac{4 + (\sin\alpha+\cos\alpha)^2}{4 – (\sin\alpha+\cos\alpha)^2} = \dfrac{4 + (1 + \sin 2\alpha)}{4 – (1 + \sin 2\alpha)} = \dfrac{5 + \sin 2\alpha}{3 – \sin 2\alpha}\)।