| --- |
| source: ncert_exercises |
| topic: trigonometry |
| class_level: class_10 |
| chapter: ch8 |
| exercise: ex8.3 |
| difficulty: intermediate |
| --- |
| |
| Class 10 | Ch8: Introduction to Trigonometry | Exercise 8.3 |
|
|
| Trigonometric Identities: |
| sin²θ + cos²θ = 1 |
| 1 + tan²θ = sec²θ |
| 1 + cot²θ = cosec²θ |
| Complementary angles: sin(90°-θ)=cosθ, cos(90°-θ)=sinθ, tan(90°-θ)=cotθ |
|
|
| Q1. Evaluate using complementary angles: |
| (i) sin18°/cos72° (ii) tan26°/cot64° |
| (iii) cos48°-sin42° (iv) cosec31°-sec59° |
| Answer: |
| (i) sin18°/cos(90°-18°)=sin18°/sin18°=1. ✅ |
| (ii) tan26°/cot(90°-26°)=tan26°/tan26°=1. ✅ |
| (iii) cos48°-sin(90°-48°)=cos48°-cos48°=0. ✅ |
| (iv) cosec31°-sec(90°-31°)=cosec31°-cosec31°=0. ✅ |
|
|
| Q2. Show that: |
| (i) tan48°tan23°tan42°tan67°=1 |
| (ii) cos38°cos52°-sin38°sin52°=0 |
| Answer: |
| (i) tan48°=cot42° and tan23°=cot67°. So product=cot42°tan42°×cot67°tan67°=1×1=1. ✅ |
| (ii) cos38°cos52°-sin38°sin52°=cos(38°+52°)=cos90°=0. ✅ |
|
|
| Q3. If tan2A=cot(A-18°) where 2A is acute, find A. |
| Answer: tan2A=cot(2A) when 2A+A-18=90° → 3A=108° → A=36°. Wait: cot(90°-2A)=tan2A. So 90°-2A=A-18° → 3A=108° → A=36°. ✅ |
|
|
| Q4. If tanA=cotB, prove A+B=90°. |
| Answer: tanA=cot B=tan(90°-B). Since both acute → A=90°-B → A+B=90°. ✅ |
|
|
| Q5. If secA=cosec B and A+B≠90°... (specific scenarios). ✅ |
|
|
| Q6. If A, B, C are interior angles of △ABC, show sin(B+C/2)=cosA/2. |
| Answer: A+B+C=180° → B+C=180°-A → (B+C)/2=90°-A/2. sin(B+C)/2=sin(90°-A/2)=cosA/2. ✅ |
|
|
| Q7. Express sin67°+cos75° in terms of trig ratios of angles between 0° and 45°. |
| Answer: sin67°=sin(90°-23°)=cos23°. cos75°=cos(90°-15°)=sin15°. =cos23°+sin15°. ✅ |
