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−112-\frac{1}{12}

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यहाँ प्रश्न 9(i) का हल दिया गया है:

  1. दिए गए त्रिकोणमितीय अनुपातों के मानों को प्रतिस्थापित करें।

    हमें ज्ञात है कि:
    sin⁡30∘=12\sin 30^\circ = \frac{1}{2}
    sec⁡30∘=23\sec 30^\circ = \frac{2}{\sqrt{3}}
    tan⁡45∘=1\tan 45^\circ = 1

  2. इन मानों को दिए गए व्यंजक में वर्ग करके रखें।

    sin⁡230∘−sec⁡230∘+tan⁡245∘\sin^2 30^\circ - \sec^2 30^\circ + \tan^2 45^\circ =(12)2−(23)2+(1)2= \left(\frac{1}{2}\right)^2 - \left(\frac{2}{\sqrt{3}}\right)^2 + (1)^2 =14−43+1= \frac{1}{4} - \frac{4}{3} + 1

  3. भिन्नों को हल करने के लिए एक सामान्य हर (denominator) ज्ञात करें।

    4 और 3 का लघुत्तम समापवर्त्य (LCM) 12 है।

    =1×34×3−4×43×4+1×121×12= \frac{1 \times 3}{4 \times 3} - \frac{4 \times 4}{3 \times 4} + \frac{1 \times 12}{1 \times 12} =312−1612+1212= \frac{3}{12} - \frac{16}{12} + \frac{12}{12}

  4. पदों को जोड़ें और घटाएँ।

    =3−16+1212= \frac{3 - 16 + 12}{12} =−13+1212= \frac{-13 + 12}{12} =−112= \boxed{-\frac{1}{12}}
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