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Class 10 ଜ୍ୟାମିତି
ତ୍ରିକୋଣମିତି Ex 4(b)

ତ୍ରିକୋଣମିତି Ex 4(b) – Study Material Class 10 ଜ୍ୟାମିତି

1. 🌟 ବିଶେଷ କୋଣର ମୂଲ୍ୟ (Values for 0°, 90°, 180°)

କୋଣ (θ\theta) sin⁡\sin cos⁡\cos tan⁡\tan cot⁡\cot sec⁡\sec csc⁡\csc
0∘0^\circ 00 11 00 - 11 -
90∘90^\circ 11 00 - 00 - 11
180∘180^\circ 00 −1-1 00 - −1-1 -

[Image of trigonometric values table]


2. 📝 ଗୁରୁତ୍ୱପୂର୍ଣ୍ଣ ସୂତ୍ରାବଳୀ (Fundamental Formulas)

ସେଟ୍ - A: ଅନୁପୂରକ ଏବଂ ସ୍ଥୂଳକୋଣ ସୂତ୍ର

  • 90∘−θ90^\circ - \theta: sin⁡(90∘−θ)=cos⁡θ\sin(90^\circ - \theta) = \cos \theta
    cos⁡(90∘−θ)=sin⁡θ\cos(90^\circ - \theta) = \sin \theta
    tan⁡(90∘−θ)=cot⁡θ\tan(90^\circ - \theta) = \cot \theta
  • 90∘+θ90^\circ + \theta: sin⁡(90∘+θ)=cos⁡θ\sin(90^\circ + \theta) = \cos \theta
    cos⁡(90∘+θ)=−sin⁡θ\cos(90^\circ + \theta) = -\sin \theta
    tan⁡(90∘+θ)=−cot⁡θ\tan(90^\circ + \theta) = -\cot \theta
  • 180∘−θ180^\circ - \theta: sin⁡(180∘−θ)=sin⁡θ\sin(180^\circ - \theta) = \sin \theta
    cos⁡(180∘−θ)=−cos⁡θ\cos(180^\circ - \theta) = -\cos \theta
    tan⁡(180∘−θ)=−tan⁡θ\tan(180^\circ - \theta) = -\tan \theta

ସେଟ୍ - B: ଯୌଗିକ କୋଣ ସୂତ୍ର (Compound Angles)

  • sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A + B) = \sin A \cos B + \cos A \sin B
  • sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin(A - B) = \sin A \cos B - \cos A \sin B
  • cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A + B) = \cos A \cos B - \sin A \sin B
  • cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A - B) = \cos A \cos B + \sin A \sin B
  • tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}
  • tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}

3. 💡 ପାଠ୍ୟପୁସ୍ତକ ସମାହିତ ଉଦାହରଣ (Book Examples)

🌟 ଉଦାହରଣ : ଯଦି AA ଓ BB ପ୍ରତ୍ୟେକ ସୂକ୍ଷ୍ମକୋଣ ଏବଂ sin⁡A=cos⁡B\sin A = \cos B ହୁଏ ତେବେ ପ୍ରମାଣ କର ଯେ, A+B=90∘A+B=90^\circ। ✅ ସମାଧାନ: BB ସୂକ୍ଷ୍ମକୋଣ ହେଲେ (90∘−B)(90^\circ - B) ମଧ୍ୟ ଏକ ସୂକ୍ଷ୍ମକୋଣ
sin⁡A=cos⁡B⇒sin⁡A=sin⁡(90∘−B)\sin A = \cos B \Rightarrow \sin A = \sin(90^\circ - B)
⇒A=90∘−B⇒A+B=90∘\Rightarrow A = 90^\circ - B \Rightarrow A + B = 90^\circ (ପ୍ରମାଣିତ)

🌟 ଉଦାହରଣ : sin⁡15∘\sin 15^\circ ର ମୂଲ୍ୟ ନିରୂପଣ କର। ✅ ସମାଧାନ: sin⁡15∘=sin⁡(45∘−30∘)\sin 15^\circ = \sin(45^\circ - 30^\circ)
=sin⁡45∘cos⁡30∘−cos⁡45∘sin⁡30∘= \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ
=(12)(32)−(12)(12)=3−122= \left( \frac{1}{\sqrt{2}} \right) \left( \frac{\sqrt{3}}{2} \right) - \left( \frac{1}{\sqrt{2}} \right) \left( \frac{1}{2} \right) = \frac{\sqrt{3} - 1}{2\sqrt{2}}

4. ❓ ଗୁରୁତ୍ୱପୂର୍ଣ୍ଣ ପ୍ରଶ୍ନବଳୀ (Important Questions)

୧. ମାନ ନିର୍ଣ୍ଣୟ କର : sin⁡18∘cos⁡72∘\frac{\sin 18^\circ}{\cos 72^\circ} ✅ ଉତ୍ତର: sin⁡18∘cos⁡(90∘−18∘)=sin⁡18∘sin⁡18∘=1\frac{\sin 18^\circ}{\cos(90^\circ - 18^\circ)} = \frac{\sin 18^\circ}{\sin 18^\circ} = 1

୨. ସରଳ କର : csc⁡31∘−sec⁡59∘\csc 31^\circ - \sec 59^\circ ✅ ଉତ୍ତର: csc⁡31∘−sec⁡(90∘−31∘)=csc⁡31∘−csc⁡31∘=0\csc 31^\circ - \sec(90^\circ - 31^\circ) = \csc 31^\circ - \csc 31^\circ = 0

୩. ପ୍ରମାଣ କର : cos⁡2(90∘+α)+cos⁡2(180∘−α)=1\cos^2(90^\circ + \alpha) + \cos^2(180^\circ - \alpha) = 1 ✅ ଉତ୍ତର: (−sin⁡α)2+(−cos⁡α)2=sin⁡2α+cos⁡2α=1(-\sin \alpha)^2 + (-\cos \alpha)^2 = \sin^2 \alpha + \cos^2 \alpha = 1

୪. ଯଦି tan⁡A=12\tan A = \frac{1}{2} ଏବଂ tan⁡B=13\tan B = \frac{1}{3}, ତେବେ A+BA+B ର ମାନ କେତେ? ✅ ଉତ୍ତର: tan⁡(A+B)=1/2+1/31−(1/2⋅1/3)=5/65/6=1⇒A+B=45∘\tan(A+B) = \frac{1/2 + 1/3}{1 - (1/2 \cdot 1/3)} = \frac{5/6}{5/6} = 1 \Rightarrow A+B = 45^\circ

୫. ମାନ ନିର୍ଣ୍ଣୟ କର : cos⁡1∘⋅cos⁡2∘⋅cos⁡3∘…cos⁡180∘\cos 1^\circ \cdot \cos 2^\circ \cdot \cos 3^\circ \dots \cos 180^\circ ✅ ଉତ୍ତର: ଏହି ଗୁଣଫଳରେ cos⁡90∘\cos 90^\circ ଅଛି, ଯାହାର ମୂଲ୍ୟ 00
ତେଣୁ ସମୁଦାୟ ଗୁଣଫଳ 00 ହେବ