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Find the value of the following: sin2 -cos2 (ii) sec2 + tan2 (iii) cos2 - sec2 (iv) tan2 - cos2 cos2 - sec2 - cos2

Mathematics

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यह रहा दूसरा प्रश्न:

  1. दिए गए त्रिकोणमितीय मानों को पहचानें।

    हमें sec⁡2π6+tan⁡2π4−sin⁡2π4\sec^2 \frac{\pi}{6} + \tan^2 \frac{\pi}{4} - \sin^2 \frac{\pi}{4} का मान ज्ञात करना है।
    यहाँ कोण हैं: π6=30∘\frac{\pi}{6} = 30^\circ और π4=45∘\frac{\pi}{4} = 45^\circ.

  2. मानक त्रिकोणमितीय मानों को याद करें।

    • sec⁡(π6)=sec⁡(30∘)=1cos⁡(30∘)=13/2=23\sec(\frac{\pi}{6}) = \sec(30^\circ) = \frac{1}{\cos(30^\circ)} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}}
    • tan⁡(π4)=tan⁡(45∘)=1\tan(\frac{\pi}{4}) = \tan(45^\circ) = 1
    • sin⁡(π4)=sin⁡(45∘)=12\sin(\frac{\pi}{4}) = \sin(45^\circ) = \frac{1}{\sqrt{2}}

  3. इन मानों को दिए गए व्यंजक में प्रतिस्थापित करें।

    sec⁡2π6+tan⁡2π4−sin⁡2π4=(23)2+(1)2−(12)2\sec^2 \frac{\pi}{6} + \tan^2 \frac{\pi}{4} - \sin^2 \frac{\pi}{4} = \left(\frac{2}{\sqrt{3}}\right)^2 + (1)^2 - \left(\frac{1}{\sqrt{2}}\right)^2

  4. व्यंजक को सरल करें।

    =43+1−12= \frac{4}{3} + 1 - \frac{1}{2}

    सामान्य हर (LCM) 6 है।

    =4×23×2+1×61×6−1×32×3= \frac{4 \times 2}{3 \times 2} + \frac{1 \times 6}{1 \times 6} - \frac{1 \times 3}{2 \times 3} =86+66−36= \frac{8}{6} + \frac{6}{6} - \frac{3}{6} =8+6−36= \frac{8 + 6 - 3}{6} =14−36= \frac{14 - 3}{6} =116= \frac{11}{6}

    अतः, व्यंजक का मान है:

    116\boxed{\frac{11}{6}}
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