---
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°. ✅