Question:
If \(\sec(\theta – \alpha)\), \(\sec\theta\) and \(\sec(\theta + \alpha)\) are in AP, where \(\cos\alpha \ne 1\), then what is the value of \(\sin^2\theta + \cos\alpha\)?
यदि \(\sec(\theta – \alpha)\), \(\sec\theta\) और \(\sec(\theta + \alpha)\) समांतर श्रेणी (AP) में हैं, जहाँ \(\cos\alpha \ne 1\), तो \(\sin^2\theta + \cos\alpha\) का मान क्या है?
Options:
- 0
- 1
- -1
- 1/2
- 0
- 1
- -1
- 1/2
Correct Answer: a)
Explanation:
Since the terms are in AP:
\[2\sec\theta = \sec(\theta – \alpha) + \sec(\theta + \alpha) = \frac{1}{\cos(\theta – \alpha)} + \frac{1}{\cos(\theta + \alpha)}\]
\[\frac{2}{\cos\theta} = \frac{2\cos\theta\cos\alpha}{\cos^2\theta – \sin^2\alpha} \implies \cos^2\theta – \sin^2\alpha = \cos^2\theta\cos\alpha\]
\[\cos^2\theta(1 – \cos\alpha) = \sin^2\alpha = (1 – \cos\alpha)(1 + \cos\alpha)\]
Since \(\cos\alpha \ne 1\), divide by \(1 – \cos\alpha\):
\[\cos^2\theta = 1 + \cos\alpha \implies 1 – \sin^2\theta = 1 + \cos\alpha \implies \sin^2\theta + \cos\alpha = 0\]
Hence, the correct option is (a).
AP की शर्त से:
\[2\sec\theta = \frac{1}{\cos(\theta – \alpha)} + \frac{1}{\cos(\theta + \alpha)} = \frac{2\cos\theta\cos\alpha}{\cos^2\theta – \sin^2\alpha}\]
\[\cos^2\theta(1 – \cos\alpha) = 1 – \cos^2\alpha = (1 – \cos\alpha)(1 + \cos\alpha)\]
\(1 – \cos\alpha\) से भाग देने पर:
\[\cos^2\theta = 1 + \cos\alpha \implies 1 – \sin^2\theta = 1 + \cos\alpha \implies \sin^2\theta + \cos\alpha = 0\]
अतः सही विकल्प (a) है।