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

এই কুইজে ত্রিকোনমিতির বিভিন্ন সূত্র এবং গণনা সম্পর্কিত প্রশ্ন রয়েছে। সাইন, কোসাইন এবং ট্যানজেন্ট এর মান নির্ণয় করার বিভিন্ন পদ্ধতি নিয়ে আলোচনা করা হয়েছে। শিক্ষার্থীদের জন্য এটি একটি মূল্যবান পরীক্ষার অংশ হতে পারে।