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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
00^\circ 00 11 00 - 11 -
9090^\circ 11 00 - 00 - 11
180180^\circ 00 1-1 00 - 1-1 -

[Image of trigonometric values table]


2. 📝 ଗୁରୁତ୍ୱପୂର୍ଣ୍ଣ ସୂତ୍ରାବଳୀ (Fundamental Formulas)

ସେଟ୍ - A: ଅନୁପୂରକ ଏବଂ ସ୍ଥୂଳକୋଣ ସୂତ୍ର

  • 90θ90^\circ - \theta: sin(90θ)=cosθ\sin(90^\circ - \theta) = \cos \theta
    cos(90θ)=sinθ\cos(90^\circ - \theta) = \sin \theta
    tan(90θ)=cotθ\tan(90^\circ - \theta) = \cot \theta
  • 90+θ90^\circ + \theta: sin(90+θ)=cosθ\sin(90^\circ + \theta) = \cos \theta
    cos(90+θ)=sinθ\cos(90^\circ + \theta) = -\sin \theta
    tan(90+θ)=cotθ\tan(90^\circ + \theta) = -\cot \theta
  • 180θ180^\circ - \theta: sin(180θ)=sinθ\sin(180^\circ - \theta) = \sin \theta
    cos(180θ)=cosθ\cos(180^\circ - \theta) = -\cos \theta
    tan(180θ)=tanθ\tan(180^\circ - \theta) = -\tan \theta

ସେଟ୍ - B: ଯୌଗିକ କୋଣ ସୂତ୍ର (Compound Angles)

  • sin(A+B)=sinAcosB+cosAsinB\sin(A + B) = \sin A \cos B + \cos A \sin B
  • sin(AB)=sinAcosBcosAsinB\sin(A - B) = \sin A \cos B - \cos A \sin B
  • cos(A+B)=cosAcosBsinAsinB\cos(A + B) = \cos A \cos B - \sin A \sin B
  • cos(AB)=cosAcosB+sinAsinB\cos(A - B) = \cos A \cos B + \sin A \sin B
  • tan(A+B)=tanA+tanB1tanAtanB\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}
  • tan(AB)=tanAtanB1+tanAtanB\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}

3. 💡 ପାଠ୍ୟପୁସ୍ତକ ସମାହିତ ଉଦାହରଣ (Book Examples)

🌟 ଉଦାହରଣ : ଯଦି AABB ପ୍ରତ୍ୟେକ ସୂକ୍ଷ୍ମକୋଣ ଏବଂ sinA=cosB\sin A = \cos B ହୁଏ ତେବେ ପ୍ରମାଣ କର ଯେ, A+B=90A+B=90^\circସମାଧାନ: BB ସୂକ୍ଷ୍ମକୋଣ ହେଲେ (90B)(90^\circ - B) ମଧ୍ୟ ଏକ ସୂକ୍ଷ୍ମକୋଣ
sinA=cosBsinA=sin(90B)\sin A = \cos B \Rightarrow \sin A = \sin(90^\circ - B)
A=90BA+B=90\Rightarrow A = 90^\circ - B \Rightarrow A + B = 90^\circ (ପ୍ରମାଣିତ)

🌟 ଉଦାହରଣ : sin15\sin 15^\circ ର ମୂଲ୍ୟ ନିରୂପଣ କର।ସମାଧାନ: sin15=sin(4530)\sin 15^\circ = \sin(45^\circ - 30^\circ)
=sin45cos30cos45sin30= \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ
=(12)(32)(12)(12)=3122= \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 : sin18cos72\frac{\sin 18^\circ}{\cos 72^\circ} ✅ ଉତ୍ତର: sin18cos(9018)=sin18sin18=1\frac{\sin 18^\circ}{\cos(90^\circ - 18^\circ)} = \frac{\sin 18^\circ}{\sin 18^\circ} = 1

୨. ସରଳ କର : csc31sec59\csc 31^\circ - \sec 59^\circ ✅ ଉତ୍ତର: csc31sec(9031)=csc31csc31=0\csc 31^\circ - \sec(90^\circ - 31^\circ) = \csc 31^\circ - \csc 31^\circ = 0

୩. ପ୍ରମାଣ କର : cos2(90+α)+cos2(180α)=1\cos^2(90^\circ + \alpha) + \cos^2(180^\circ - \alpha) = 1 ✅ ଉତ୍ତର: (sinα)2+(cosα)2=sin2α+cos2α=1(-\sin \alpha)^2 + (-\cos \alpha)^2 = \sin^2 \alpha + \cos^2 \alpha = 1

୪. ଯଦି tanA=12\tan A = \frac{1}{2} ଏବଂ tanB=13\tan B = \frac{1}{3}, ତେବେ A+BA+B ର ମାନ କେତେ? ✅ ଉତ୍ତର: tan(A+B)=1/2+1/31(1/21/3)=5/65/6=1A+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 : cos1cos2cos3cos180\cos 1^\circ \cdot \cos 2^\circ \cdot \cos 3^\circ \dots \cos 180^\circ ✅ ଉତ୍ତର: ଏହି ଗୁଣଫଳରେ cos90\cos 90^\circ ଅଛି, ଯାହାର ମୂଲ୍ୟ 00
ତେଣୁ ସମୁଦାୟ ଗୁଣଫଳ 00 ହେବ