Question:
If $(\sin x + \sin y) = a$ and $(\cos x + \cos y) = b$, then find the value of $(\sin x \sin y + \cos x \cos y)$.
यदि $(\sin x + \sin y) = a$ और $(\cos x + \cos y) = b$ है, तो $(\sin x \sin y + \cos x \cos y)$ का मान ज्ञात कीजिए।
Options:
- a) (a^2 + b^2 + 2)/2
- b) (a^2 – b^2 – 2)/2
- c) (a^2 + b^2 – 2)/2
- d) (a^2 + b^2 – 1)/2
- a) (a^2 + b^2 + 2)/2
- b) (a^2 – b^2 – 2)/2
- c) (a^2 + b^2 – 2)/2
- d) (a^2 + b^2 – 1)/2
correct answer : c)
Explanation:
Given:
1. $\sin x + \sin y = a$
2. $\cos x + \cos y = b$
Square both equations:
$(\sin x + \sin y)^2 = a^2 \implies \sin^2 x + \sin^2 y + 2\sin x \sin y = a^2$
$(\cos x + \cos y)^2 = b^2 \implies \cos^2 x + \cos^2 y + 2\cos x \cos y = b^2$
Now, add the two squared equations:
$(\sin^2 x + \cos^2 x) + (\sin^2 y + \cos^2 y) + 2(\sin x \sin y + \cos x \cos y) = a^2 + b^2$
Using the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$:
$1 + 1 + 2(\sin x \sin y + \cos x \cos y) = a^2 + b^2$
$\implies 2 + 2(\sin x \sin y + \cos x \cos y) = a^2 + b^2$
$\implies 2(\sin x \sin y + \cos x \cos y) = a^2 + b^2 – 2$
$\implies \sin x \sin y + \cos x \cos y = \frac{a^2 + b^2 – 2}{2}$.
Hence, option (c) is the correct answer.
दिया गया है:
1. $\sin x + \sin y = a$
2. $\cos x + \cos y = b$
दोनों समीकरणों का वर्ग करें:
$(\sin x + \sin y)^2 = a^2 \implies \sin^2 x + \sin^2 y + 2\sin x \sin y = a^2$
$(\cos x + \cos y)^2 = b^2 \implies \cos^2 x + \cos^2 y + 2\cos x \cos y = b^2$
अब, दोनों वर्ग किए गए समीकरणों को जोड़ें:
$(\sin^2 x + \cos^2 x) + (\sin^2 y + \cos^2 y) + 2(\sin x \sin y + \cos x \cos y) = a^2 + b^2$
त्रिकोणमितीय सर्वसमिका $\sin^2 \theta + \cos^2 \theta = 1$ का उपयोग करते हुए:
$1 + 1 + 2(\sin x \sin y + \cos x \cos y) = a^2 + b^2$
$\implies 2 + 2(\sin x \sin y + \cos x \cos y) = a^2 + b^2$
$\implies 2(\sin x \sin y + \cos x \cos y) = a^2 + b^2 – 2$
$\implies \sin x \sin y + \cos x \cos y = \frac{a^2 + b^2 – 2}{2}$।
अतः, विकल्प (c) सही उत्तर है।