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Find the value of the following: ① sin²π/4 - cos²π/4 ② sec²π/6 + csc²π/3 ③ cos²π/4 + sec²π/3 + tan²π/6 ④ sec²π/6 - sec²π/4 ⑤ tan²π/3 - cosec²π/6 + sec²π/6

Mathematics

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Answer

−56-\frac{5}{6}

Step-by-step solution

4 steps

यह रहा पहला प्रश्न:

  1. दिए गए त्रिकोणमितीय मानों को पहचानें।

    हमें sin⁡2π4−cot⁡2π3−csc⁡2π2\sin^2 \frac{\pi}{4} - \cot^2 \frac{\pi}{3} - \csc^2 \frac{\pi}{2} का मान ज्ञात करना है।
    यहाँ कोण हैं: π4=45∘\frac{\pi}{4} = 45^\circ, π3=60∘\frac{\pi}{3} = 60^\circ, और π2=90∘\frac{\pi}{2} = 90^\circ.

  2. मानक त्रिकोणमितीय मानों को याद करें।

    • sin⁡(π4)=sin⁡(45∘)=12\sin(\frac{\pi}{4}) = \sin(45^\circ) = \frac{1}{\sqrt{2}}
    • cot⁡(π3)=cot⁡(60∘)=13\cot(\frac{\pi}{3}) = \cot(60^\circ) = \frac{1}{\sqrt{3}}
    • csc⁡(π2)=csc⁡(90∘)=1sin⁡(90∘)=11=1\csc(\frac{\pi}{2}) = \csc(90^\circ) = \frac{1}{\sin(90^\circ)} = \frac{1}{1} = 1
  3. इन मानों को दिए गए व्यंजक में प्रतिस्थापित करें।

    sin⁡2π4−cot⁡2π3−csc⁡2π2=(12)2−(13)2−(1)2\sin^2 \frac{\pi}{4} - \cot^2 \frac{\pi}{3} - \csc^2 \frac{\pi}{2} = \left(\frac{1}{\sqrt{2}}\right)^2 - \left(\frac{1}{\sqrt{3}}\right)^2 - (1)^2
  4. व्यंजक को सरल करें।

    =12−13−1= \frac{1}{2} - \frac{1}{3} - 1

    सामान्य हर (LCM) 6 है।

    =36−26−66= \frac{3}{6} - \frac{2}{6} - \frac{6}{6} =3−2−66= \frac{3 - 2 - 6}{6} =1−66= \frac{1 - 6}{6} =−56= \frac{-5}{6}

    अतः, व्यंजक का मान है:

    −56-\frac{5}{6}
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