Trigonometry Quiz for Class 10

ত্রিকোণমিতিক অনুপাত

• tan(-1590°) = tan(1590°) = tan(4.360°+150°) = -tan150° = tan(180°-30°) = + tan30° = √3/ 3
• sin (-1230°) - cos {(2n + 1) (π/3)} = sin 1230° - cos 2n +π/3 = sin (3.360°+ 150°) - cos (π/3) = -sin 150°-(-cos(π/3)) = sin (180° -30°) + cos(π/3) = sin 30°+ cos(π/3) = 1/2 + 1/2= 1
• sin 780° cos 390°+ sin (330°) cos( - 300°) = sin(2.360°+ 60°) cos (360°+30°) = sin780° cos 390°-sin330° cos 300° = sin (360°-30°) cos (360°~60°) = sin60°cos30°-(-sin30°)cos60° = √3/2 * √3/2 - (-1/2)*1/2 = 3/4 + 1/4 = 1
• cos420°sin(-300°) – sin870°cos570° =cos420°(-sin300°)-sin 870°cos570° = cos (360°+60°) sin(360°~60°) -sin(2.360°+150°) cos(2.360°-150°) = -cos60°(-sin60°)- sin150°cos150° = cos 60° sin 60° − sin (180°− 30°) cos (180°-30°) = cos 60° sin 60°- sin 30° (− cos 30°) = 1/2 * √3/2 - 1/2 * (-√3/2) = √3/4 + √3/4 = √3/2

প্রশ্নমালা VII A

• sin² (25π/14) +sin2 (28π/14)+ sin² (29π/14) = sin² (π/7) + sin² (2π/7) + sin² (3π/7) = sin² (π/7) + cos² (π/7) + sin² (π/7) + cos² (π/7) = 2 (sin² (π/7) + cos² (π/7)) = 2.1 = 2
• sin² (25π/18) + sin² (237π/18)+cos² (23π/8)+cos² (2π/8) = sin² 2 (π/18) + sin² (3π/18) + cos² (2π/8) + cos² (23π/8) = sin² (π/18) + sin² (π/6) + cos² (π/6) + cos² (π/4) = (sin² (π/18) + cos² (π/18) + sin² (π/6) + cos² (π/6) + cos² (π/4) = 1+1=2
• sec² (14π/17) -sec² (39π/17) = sec 2(- (3π/17)) - sec² (2π/17)+ cot (2π/17)- cot 2 (π/17 - 5/17) = sec² (3π/17)- sec² (2π/17)+ cot (2π/17)- cot 2 (11π/34) = sec² (3π/17)- sec² (2π/17)+ cot (2π/17)- cot 2 (27π/34) = sec² (3π/17)- sec² (2π/17) + cot (5π/17) - cot (2π/17)- cot (11π/34) = sec² (3π/17)- sec² (2π/17) + cot 2 (5π/17) - sec² (5π/17) + cot (3π/17) =(sec² (3π/17)- sec² (5π/17))-(sec² (3π/17)- tan 2 (5π/17) = seс 2( 3π/17 ) - sec ² (5π/17 ) - cot (3π/17) = seс 2 ( 3π/17 ) - tan 2 ( 3π/17 ) = sec² (3π/17) - tan² (3π/17) = 1
• sin 2 (π/12) এর মান নির্ণয় কর। = tan15° + tan 45°+ tan(90°-15°)+ tan (90° +15°)+tan (180-45°) + tan (180°-15°) = tan15°+ tan 45° + cot15°-cot15°- tan45°-tan15°=0
• -1-1=0 (Ans.)
• cos2 (π/24) +cos2 (2π/24)+cos2 (19π/24)+cos2 (31π/24)+cos2 (19π/24)+cos2 (219π/24)+cos2 (37π/24) = cos2 (π/24) +cos2 (2π/24)+cos2 (19π/24)+cos2 (31π/24)+cos2(π/24)+cos2 (19π/24)+cos2 (37π/24) = (sin² (19π/24) +cos² (19π/24) + sin² (19π/24) + cos² (19π/24))+ (sin² (19π/24) +cos² (19π/24) + sin² (19π/24) + cos² (19π/24))+ (sin² (19π/24)+cos² (19π/24)+sin² (37π/24)+cos² (37π/24)) = 1 + 1 = 2 (Ans.)
• sin² (23π/12) + sin² (25π/12) + sin² (29π/12) + sin² (211π/12) +sin2 (3π/12)+ sin2 (9π/12)+ sin2 (5π/12)+ sin2 (11π/12) =sin2 (π/12) +sin2 (3π/12) + sin2 (5π/12) + sin2 (2π/12) + sin 2 (3π/12) +sin2 (5π/12) +cos2 (π/12) + cos2 (π/3) =(sin² (π/12) + cos² (π/12)) + (sin² (π/3) + cos² (π/3) + (sin 2 (5π/12) + cos² (5π/12) =1+1+1=3(Ans.)
• tan15°+tan45°+ tan75°+……+ tan165° = tan15°+ tan 45°+ tan 75°+ tan105°+ tan135°+tan165° = tan15°+ tan 45° + cot15°-cot15°- tan45°-tan15°=0(Ans.)
• cos² 150+ cos² 225° -- cos 235°+.+ cos² 75° =cos²15°+cos² 25°+ cos² 35° + cos² 45° + cos² 55°+ cos² 65° + cos² 75 = cos²15° + cos² 25°+ cos 235° +()² √2 + cos² (90°-35°) + cos ² (90° − 25°) +cos² (90°-15°) 2 = cos 215°+ cos ¡² 25° + cos ² 35°+¦ + 2 = sin 235°+ sin225°+ sin 215° 25π 27π + sin² 12 12 12 + sin 2 9π + sin^2 11π =sin2 12 12 12 2 3π 25π + sin² + sin 12 12 12 3π 5π +cos2 + cos2 12 12 π =(sin²+ cos²) + (sin² 12 + (sin 2 5π 12 12 + cos =1+1+1=3(Ans.)
• cos 2 25°+ cos ² 35°+ cos ² 45° + cos 255°+ cos ² 65° = cos 225° + cos 235° + 2()²+ cos² (90° -35°) + cos² (90° - 25°) = cos²25° + cos² 35° + 1 + sin 235° 2 + sin 225° = (sin 225° + cos $ 225°) + + + (sin225° + cos225°) 1+1 = 1 + 2 = 5 2 (Ans.)
• sin 2 10° + sin 220° + sin230°+.....+sin280° = sin 210°+ sin 220°+ sin 230° + sin 240°+ sin 250° + sin ² 60° + sin 270° + sin280° = sin 210°+ sin 220°+sin230°+ sin240°+sin2(90°-40°) + sin²(90° – 30°) + sin ² (90° -20°) + sin²(90°-10°) = sin 2 10°+ sin 220°+ sin 230° + sin 240°+ cos 240°+ cos 230° + cos 220° + cos 210° = (sin²10° + cos210°) + (sin² 20° + cos²20°) + (sin230° + cos²30°) + (sin240°+ cos² 40°) = 1+1+1+1 = 4 (Ans.)

উচ্চতর গণিত: ১ম পত্র সমাধান

• tane + sec(-0) = cote + cosec(-0) = -5-13 = -12 13 = 12 -13 5 = 5 = (--)x(-)=-x-= 3 531 25 2 5 10 5 18. 12
• sin e = এবং 90° <<180° হলে, tan 0+ sec (10) = cot0+ cos ec(-0) = 10 3 = 5 12 = 12 13 = 1 144 25 = 1 169 169 13 13 => 5 12 => 13 5 => 12 => 13 = -5-13 = 13 12 12 =5x= 5 18 3 x(-) = 12 12 3 12
• cote = => cote = এবং cos e ঋণাত্মক হলে, cot (0) + cos eco = cos 0+ sin (-0) = 3 5 4 = 3 5 4 = 16 1+ 9 25 5 9 = -২ এবং 5 = tane cose = 3.x(-) 5.= 4 5 = 54 = 4 = -3-5 5 = X 3-4 4 -3+4 = 55 40 = -10 (Ans.)

যোগফল নির্ণয়

• sinx + sin(π + x) + sin(2π + x) + ......(n+1)তম পদ পর্যন্ত = sinx - sinx + sinx - sinx + (n+1) তম পদ পর্যন্ত = 0 (যখন n বিজোড় ) = sinx (যখন n জোড়)
• cos e + cos (π + θ) + cos (2π + 0) + ...+ cos (η π + 0) = cos + {-cose + cose cose + + (-1)" cos e } = cose (যখন n জোড়) = 0 (যখন n বিজোড়)
• tane + tan(π + Θ) + tan(2 π + 0) + + tan(η π + 0) = tane + tane + tane + ...+ (n + 1) তম পদ পর্যন্ত = (n + 1) tane (Ans.)
• n ∈ Z হলে, sin{nπ+ (−1)".π/4 = sin{n+ (−1)"} এর মান নির্ণয় কর। sin{nx + (-1)"} π/4 = sin {2m + (−1)2} π/4 = sin (2m + 1) π/4 = sin π/4 = 1 (যখন n জোড়) = sin (2m + 1) + (−1)2m+1π/4 π/4 = sin{2mπ + (π−1)} π/4 = sin(π - π/4) = sin π/4 = 1 (যখন n বিজোড়) = -1 (Ans.)

প্রশ্নমালা-VII B

• cosec 165° = cosec (90° +75°) = sec 75° = 1 cos.75° = 1 cos(45° +30°) = 1 2√2 = √3-1 2√2(√3+1)_2(√6+√2) = (√3−1)(√3+1) 2(√6 + √2) =√6+ √2 2
• cos 38°15′ sin 68°15′ – cos51°45′sin21°45′ = cos 38°15′ sin 68°15′ – cos (90°-38°15′) sin (90°- 68°15′) = cos 38°15′ sin 68°15′-.sin 38°15′ cos 68°15′ = sin (68°15′-38°15′) = sin30°=- 1 2
• cos 69°22′ cos 9°22′ + cos 80°38′ cos 20°38′ = cos 69°22′ cos 9°22′ + cos (90° 9°22') cos (90° – 69°22′) | = cos 69°22′ cos 9°22′ + = cos (69°22′- 9°22′) = cos 60° = 1 2
• sin 76°40′cos 16°40′- cos.73°20′sin 13°20′ =sin76°40′cos16°40′- cos(90°-16°40′) sin (90°- 76°40′) = sin 76°40′cos 16°40′- sin16°40′cos 76°40′ = sin (76°40′ -16°40′ )= sin60°= √3/2
• cos 17°40′ sin 77°40′ + cos 107°40′ sin 12°20′ = cos 17°40′ sin 77°40′ + .cos (90° +17°40′) sin(90°-77°40′) = cos17°40′sin77°40′- sin17°40′cos77°40′ = sin (77°40′-17°40′) = sin60°= √3/2
• cos 74°33' cos 14°33'+ cos 75°27' cos 15°27' = cos 74°33' cos 14°33'+ cos (90°-14°33') cos (90°-74°33') = cos 69°22′ cos 9°22′+ sin 14°33' sin 74°33''' = cos (74°33'-14°33') = cos 60° = 1/2