Question:
It is given that \(\cos(\theta – \alpha) = a\), \(\cos(\theta – \beta) = b\). What is \(\cos(\alpha – \beta)\) equal to?
दिया गया है कि \(\cos(\theta – \alpha) = a\) और \(\cos(\theta – \beta) = b\) है। \(\cos(\alpha – \beta)\) किसके बराबर है?
Options:
- \(a\sqrt{1-b^2} + b\sqrt{1-a^2}\)
- \(ab + \sqrt{1-a^2}\sqrt{1-b^2}\)
- \(ab – \sqrt{1-a^2}\sqrt{1-b^2}\)
- \(a\sqrt{1-a^2} – b\sqrt{1-a^2}\)
- \(a\sqrt{1-b^2} + b\sqrt{1-a^2}\)
- \(ab + \sqrt{1-a^2}\sqrt{1-b^2}\)
- \(ab – \sqrt{1-a^2}\sqrt{1-b^2}\)
- \(a\sqrt{1-a^2} – b\sqrt{1-a^2}\)
Correct Answer: b)
Explanation:
\(\cos(\alpha – \beta) = \cos((\theta – \beta) – (\theta – \alpha)) = \cos(\theta-\beta)\cos(\theta-\alpha) + \sin(\theta-\beta)\sin(\theta-\alpha)\).
Given \(\cos(\theta-\alpha) = a \implies \sin(\theta-\alpha) = \sqrt{1-a^2}\) and \(\cos(\theta-\beta) = b \implies \sin(\theta-\beta) = \sqrt{1-b^2}\).
\(\cos(\alpha-\beta) = ab + \sqrt{1-a^2}\sqrt{1-b^2}\).
\(\cos(\alpha – \beta) = \cos((\theta – \beta) – (\theta – \alpha)) = \cos(\theta-\beta)\cos(\theta-\alpha) + \sin(\theta-\beta)\sin(\theta-\alpha)\)।
\(= ab + \sqrt{1-a^2}\sqrt{1-b^2}\)।