NCERT Solutions for Class 10 Maths Chapter 8
Important NCERT Questions
Introduction Trigonometry
NCERT Books for Session 2022-2023
CBSE Board and UP Board Others state Board
EXERCISE 8.2
Page No:187
Questions No: 1
Evaluate the following :
(i) sin 60° cos 30° + sin 30° cos 60°
(ii) 2 tan2 45° + cos2 30° – sin2 60°
(iii) cos 45° / sec 30° + cosec 30°
(iv) sin 30° + tan 45° – cosec 60° / sec 30° + cos 60° + cot 45°
(v) 5 cos² 60° + 4 sec² 30° – tan² 45°/ sin² 30° + cos² 30°
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(i) sin 60° cos 30° + sin 30° cos 60°
Putting the value of each trigonometric ratio, we get
(√3/2)(√3/2)+(1/2)(1/2) = 3/4 + 1/4 = 4/4 = 1
(ii) Putting the value of each trigonometric ratio, we get
2(1)²+(√3/2)²-(√3/2)² = 2 + 3/4 – 3/4 = 1
(iii) (cos 45°)/(sec 30°+cosec 30°)
Putting the value of each trigonometric ratios, we get
(1/√2)/(2/√3 + 2) = (1/√2)/((2+2√3)/√3) = √3/(√2(2+2√3)) = √3/(2√2 + 2√6)
= √3/(2√2 + 2√6) × (2√2 – 2√6)/(2√2 – 2√6) = (2√6 – 2√18)/((2√2)² – (2√6)²) = (2√6 – 6√2)/(8 – 24) = (-2(3√2 – √6))/(-16) = (3√2 – √6)/(8)
(iv) (sin 30° + tan 45° – cosec 60°)/(sec 30° + cos 60° + cot 45°)
Putting the value of each trigonometric ratios, we get
(1/2 + 1 – 2√3)/(2√3 + 1/2 + 1) = ((√3 + 2√3 – 4)/(2√3))/((4 + √3 + 2√3)/(2√3)) = (3√3 – 4)/(3√3 + 4)
= (3√3 – 4)/(3√3 + 4) × (3√3 – 4)/(3√3 – 4) = (27 – 12√3 – 12√3 + 16)/((3√3)² – 4²)
= (43 – 24√3)/(27 – 16) = (43 – 24√3)/(11)
(v) (5 cos²60° + 4sec²30° – tan²45°)/(sin²30° + cos²30°)
Putting the value of each trigonometric ratios, we get
(5(1/2)² + 4(2/√3)² – (1)²)/((1/2)² + (√3/2)²) = (5/4 + 16/3 – 1)/(1/4 + 3/4) = ((15 + 64 -12)/(12))/(4/4) = 67/12
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