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निम्न तीन (03) प्रश्नों के लिए निम्नलिखित पर विचार कीजिए: माना कि , =sin25∘ और है।
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दिया गया है,
सरल करने के लिए व्यंजक pq + qr + rp है, जहाँ:
\(p = \sin(35^\circ) \)
\(q = \sin(25^\circ) \)
\(r = \sin(-95^\circ) = -\cos(5^\circ) \)
इस प्रकार, हमें निम्न की गणना करनी है:
\( pq + qr + rp = \sin(35^\circ) \sin(25^\circ) + \sin(25^\circ) (-\cos(5^\circ)) + (-\cos(5^\circ)) \sin(35^\circ) \)
त्रिकोणमितीय सर्वसमिकाओं का उपयोग करके, हम व्यंजक को सरल करते हैं:
\( pq + qr + rp = -\sin^2(25^\circ) - \sin(95^\circ) \sin(35^\circ) \)
योग-से-गुणफल सर्वसमिका और ज्या और कोज्या के ज्ञात मानों का उपयोग करने पर:
\( = \frac{1}{2} [-2 \sin(25^\circ) - 2 \sin(95^\circ) \cdot \sin(35^\circ)] \)
\( = \frac{1}{2} [-1 + \cos(50^\circ) + \cos(130^\circ) - \cos(60^\circ)] \)
\( = \frac{1}{2} [-1 + \cos(50^\circ) - \cos(50^\circ) - \frac{1}{2}] \)
\( = \frac{1}{2} \left[ -1 - \frac{1}{2} \right] = -\frac{3}{4} \)
∴ pq + qr + rp का मान \( -\frac{3}{4} .\) है।
इसलिए, सही उत्तर विकल्प 1 है।
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