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Page 1 JEE Mains Previous Year Questions (2021-2024): Trigonometric Ratios, Functions & Equations 2024 Q1 - 2024 (01 Feb Shift 1) If t a n ? A = 1 v ?? ( ?? 2 + ?? + 1 ) , t a n ? ?? = v ?? v ?? 2 + ?? + 1 and t a n ? ?? = ( ?? - 3 + ?? - 2 + ?? - 1 ) 1 2 , 0 < ?? , ?? , ?? < ?? 2 , then A + B is equal to : (1) C (2) ?? - ?? (3) 2 ?? - ?? (4) ?? 2 - ?? Q2 - 2024 (30 Jan Shift 2) For ?? , ?? ? ( 0 , ?? 2 ), let 3 s i n ? ( ?? + ?? ) = 2 s i n ? ( ?? - ?? ) and a real number k be such that t a n ? ?? = kt a n ? ?? . Then the value of ?? is equal to : (1) - 2 3 (2) -5 (3) 2 3 (4) 5 Q3 - 2024 (01 Feb Shift 2) The number of solutions of the equation 4 sin 2 ? ?? - 4 c o s 3 ? ?? + 9 - 4 c o s ? ?? = 0 ; ?? ? [ - 2 ?? , 2 ?? ] is : (1) 1 (2) 3 (3) 2 Page 2 JEE Mains Previous Year Questions (2021-2024): Trigonometric Ratios, Functions & Equations 2024 Q1 - 2024 (01 Feb Shift 1) If t a n ? A = 1 v ?? ( ?? 2 + ?? + 1 ) , t a n ? ?? = v ?? v ?? 2 + ?? + 1 and t a n ? ?? = ( ?? - 3 + ?? - 2 + ?? - 1 ) 1 2 , 0 < ?? , ?? , ?? < ?? 2 , then A + B is equal to : (1) C (2) ?? - ?? (3) 2 ?? - ?? (4) ?? 2 - ?? Q2 - 2024 (30 Jan Shift 2) For ?? , ?? ? ( 0 , ?? 2 ), let 3 s i n ? ( ?? + ?? ) = 2 s i n ? ( ?? - ?? ) and a real number k be such that t a n ? ?? = kt a n ? ?? . Then the value of ?? is equal to : (1) - 2 3 (2) -5 (3) 2 3 (4) 5 Q3 - 2024 (01 Feb Shift 2) The number of solutions of the equation 4 sin 2 ? ?? - 4 c o s 3 ? ?? + 9 - 4 c o s ? ?? = 0 ; ?? ? [ - 2 ?? , 2 ?? ] is : (1) 1 (2) 3 (3) 2 (4) 0 Q4 - 2024 (27 Jan Shift 1) Let the set of all ?? ? ?? such that the equation c o s ? 2 ?? + ?? sin ? ?? = 2 ?? - 7 has a solution be [ ?? , ?? ] and ?? = t a n ? 9 ° - t a n ? 27 ° - 1 c o t ? 63 ° + t a n ? 81 ° , then pqr is equal to Q5 - 2024 (27 Jan Shift 2) If 2 t a n 2 ? ?? - 5 s e c ? ?? = 1 has exactly 7 solutions in the interval [ 0 , ???? 2 ], for the least value of ?? ? ?? then ? k = 1 n ? k 2 k is equal to : (1) 1 2 15 ( 2 14 - 14 ) (2) 1 2 14 ( 2 15 - 15 ) (3) 1 - 15 2 13 (4) 1 2 13 ( 2 14 - 15 ) Q6 - 2024 (29 Jan Shift 1) If ?? , - ?? 2 < ?? < ?? 2 is the solution of 4 c o s ? ?? + 5 s i n ? ?? = 1, then the value of t a n ? ?? is (1) 10 - v 10 6 (2) 10 - v 10 12 (3) v 10 - 10 12 (4) v 10 - 10 6 Q7- 2024 (29 Jan Shift 2) The sum of the solutions ?? ? R of the equation 3 c o s ? 2 ?? + c o s 3 ? 2 ?? c o s 6 ? ?? - s i n 6 ? ?? = ?? 3 - ?? 2 + 6 is (1) 0 (2) 1 Page 3 JEE Mains Previous Year Questions (2021-2024): Trigonometric Ratios, Functions & Equations 2024 Q1 - 2024 (01 Feb Shift 1) If t a n ? A = 1 v ?? ( ?? 2 + ?? + 1 ) , t a n ? ?? = v ?? v ?? 2 + ?? + 1 and t a n ? ?? = ( ?? - 3 + ?? - 2 + ?? - 1 ) 1 2 , 0 < ?? , ?? , ?? < ?? 2 , then A + B is equal to : (1) C (2) ?? - ?? (3) 2 ?? - ?? (4) ?? 2 - ?? Q2 - 2024 (30 Jan Shift 2) For ?? , ?? ? ( 0 , ?? 2 ), let 3 s i n ? ( ?? + ?? ) = 2 s i n ? ( ?? - ?? ) and a real number k be such that t a n ? ?? = kt a n ? ?? . Then the value of ?? is equal to : (1) - 2 3 (2) -5 (3) 2 3 (4) 5 Q3 - 2024 (01 Feb Shift 2) The number of solutions of the equation 4 sin 2 ? ?? - 4 c o s 3 ? ?? + 9 - 4 c o s ? ?? = 0 ; ?? ? [ - 2 ?? , 2 ?? ] is : (1) 1 (2) 3 (3) 2 (4) 0 Q4 - 2024 (27 Jan Shift 1) Let the set of all ?? ? ?? such that the equation c o s ? 2 ?? + ?? sin ? ?? = 2 ?? - 7 has a solution be [ ?? , ?? ] and ?? = t a n ? 9 ° - t a n ? 27 ° - 1 c o t ? 63 ° + t a n ? 81 ° , then pqr is equal to Q5 - 2024 (27 Jan Shift 2) If 2 t a n 2 ? ?? - 5 s e c ? ?? = 1 has exactly 7 solutions in the interval [ 0 , ???? 2 ], for the least value of ?? ? ?? then ? k = 1 n ? k 2 k is equal to : (1) 1 2 15 ( 2 14 - 14 ) (2) 1 2 14 ( 2 15 - 15 ) (3) 1 - 15 2 13 (4) 1 2 13 ( 2 14 - 15 ) Q6 - 2024 (29 Jan Shift 1) If ?? , - ?? 2 < ?? < ?? 2 is the solution of 4 c o s ? ?? + 5 s i n ? ?? = 1, then the value of t a n ? ?? is (1) 10 - v 10 6 (2) 10 - v 10 12 (3) v 10 - 10 12 (4) v 10 - 10 6 Q7- 2024 (29 Jan Shift 2) The sum of the solutions ?? ? R of the equation 3 c o s ? 2 ?? + c o s 3 ? 2 ?? c o s 6 ? ?? - s i n 6 ? ?? = ?? 3 - ?? 2 + 6 is (1) 0 (2) 1 (3) -1 (4) 3 Q8 - 2024 (30 Jan Shift 1) If 2 sin 3 ? ?? + sin ? 2 ?? c o s ? ?? + 4 s i n ? ?? - 4 = 0 has exactly 3 solutions in the interval [ 0 , n ?? 2 ] , n ? N, then the roots of the equation ?? 2 + ???? + ( ?? - 3 ) = 0 belong to : (1) ( 0 , 8 ) (2) ( - 8 , 0 ) (3) ( - v 17 2 , v 17 2 ) (4) Z Answer Key Q1 (1) Q2 (2) Q3 (4) Q4 (48) Q5 (4) Q6 (3) Q7 (3) Q8 (2) Solutions Q1 Finding t a n ? ( ?? + ?? ) we get Page 4 JEE Mains Previous Year Questions (2021-2024): Trigonometric Ratios, Functions & Equations 2024 Q1 - 2024 (01 Feb Shift 1) If t a n ? A = 1 v ?? ( ?? 2 + ?? + 1 ) , t a n ? ?? = v ?? v ?? 2 + ?? + 1 and t a n ? ?? = ( ?? - 3 + ?? - 2 + ?? - 1 ) 1 2 , 0 < ?? , ?? , ?? < ?? 2 , then A + B is equal to : (1) C (2) ?? - ?? (3) 2 ?? - ?? (4) ?? 2 - ?? Q2 - 2024 (30 Jan Shift 2) For ?? , ?? ? ( 0 , ?? 2 ), let 3 s i n ? ( ?? + ?? ) = 2 s i n ? ( ?? - ?? ) and a real number k be such that t a n ? ?? = kt a n ? ?? . Then the value of ?? is equal to : (1) - 2 3 (2) -5 (3) 2 3 (4) 5 Q3 - 2024 (01 Feb Shift 2) The number of solutions of the equation 4 sin 2 ? ?? - 4 c o s 3 ? ?? + 9 - 4 c o s ? ?? = 0 ; ?? ? [ - 2 ?? , 2 ?? ] is : (1) 1 (2) 3 (3) 2 (4) 0 Q4 - 2024 (27 Jan Shift 1) Let the set of all ?? ? ?? such that the equation c o s ? 2 ?? + ?? sin ? ?? = 2 ?? - 7 has a solution be [ ?? , ?? ] and ?? = t a n ? 9 ° - t a n ? 27 ° - 1 c o t ? 63 ° + t a n ? 81 ° , then pqr is equal to Q5 - 2024 (27 Jan Shift 2) If 2 t a n 2 ? ?? - 5 s e c ? ?? = 1 has exactly 7 solutions in the interval [ 0 , ???? 2 ], for the least value of ?? ? ?? then ? k = 1 n ? k 2 k is equal to : (1) 1 2 15 ( 2 14 - 14 ) (2) 1 2 14 ( 2 15 - 15 ) (3) 1 - 15 2 13 (4) 1 2 13 ( 2 14 - 15 ) Q6 - 2024 (29 Jan Shift 1) If ?? , - ?? 2 < ?? < ?? 2 is the solution of 4 c o s ? ?? + 5 s i n ? ?? = 1, then the value of t a n ? ?? is (1) 10 - v 10 6 (2) 10 - v 10 12 (3) v 10 - 10 12 (4) v 10 - 10 6 Q7- 2024 (29 Jan Shift 2) The sum of the solutions ?? ? R of the equation 3 c o s ? 2 ?? + c o s 3 ? 2 ?? c o s 6 ? ?? - s i n 6 ? ?? = ?? 3 - ?? 2 + 6 is (1) 0 (2) 1 (3) -1 (4) 3 Q8 - 2024 (30 Jan Shift 1) If 2 sin 3 ? ?? + sin ? 2 ?? c o s ? ?? + 4 s i n ? ?? - 4 = 0 has exactly 3 solutions in the interval [ 0 , n ?? 2 ] , n ? N, then the roots of the equation ?? 2 + ???? + ( ?? - 3 ) = 0 belong to : (1) ( 0 , 8 ) (2) ( - 8 , 0 ) (3) ( - v 17 2 , v 17 2 ) (4) Z Answer Key Q1 (1) Q2 (2) Q3 (4) Q4 (48) Q5 (4) Q6 (3) Q7 (3) Q8 (2) Solutions Q1 Finding t a n ? ( ?? + ?? ) we get ? ? t a n ? ( A + B ) = t a n ? ?? + t a n ? ?? 1 - t a n ? ?? t a n ? ?? = 1 v ?? ( ?? 2 + ?? + 1 ) + v ?? v ?? 2 + ?? + 1 1 - 1 ?? 2 + ?? + 1 ? ? t a n ? ( A + B ) = ( 1 + ?? ) ( v ?? 2 + ?? + 1 ) ( ?? 2 + ?? ) ( v ?? ) ( 1 + ?? ) ( v ?? 2 + ?? + 1 ) ( ?? 2 + ?? ) ( v ?? ) t a n ? ( ?? + ?? ) = v ?? 2 + ?? + 1 ?? v ?? = t a n ? ?? ?? + ?? = ?? Q2 3 s i n ? ?? c o s ? ?? + 3 s i n ? ?? c o s ? ?? = 2 s i n ? ?? c o s ? ?? - 2 s i n ? ?? c o s ? ?? 5 s i n ? ?? c o s ? ?? = - sin ? ?? c o s ? ?? t a n ? ?? = - 1 5 t a n ? ?? t a n ? ?? = - 5 t a n ? ?? Q3 4 sin 2 ? ?? - 4 c o s 3 ? ?? + 9 - 4 c o s ? ?? = 0 ; ?? ? [ - 2 ?? , 2 ?? ] 4 - 4 c o s 2 ? ?? - 4 c o s 3 ? ?? + 9 - 4 c o s ? ?? = 0 4 c o s 3 ? ?? + 4 c o s 2 ? ?? + 4 c o s ? ?? - 13 = 0 4 c o s 3 ? ?? + 4 c o s 2 ? ?? + 4 c o s ? ?? = 13 L.H.S. = 12 c a n ' t be equal to 13 . Q4 c o s ? 2 ?? + ?? · sin ? ?? = 2 ?? - 7 ?? ( sin ? ?? - 2 ) = 2 ( sin ? ?? - 2 ) ( sin ? ?? + 2 ) sin ? ?? = 2 , ? ?? = 2 ( sin ? ?? + 2 ) ? ?? ? [ 2 , 6 ] ?? = 2 ? ?? = 6 Page 5 JEE Mains Previous Year Questions (2021-2024): Trigonometric Ratios, Functions & Equations 2024 Q1 - 2024 (01 Feb Shift 1) If t a n ? A = 1 v ?? ( ?? 2 + ?? + 1 ) , t a n ? ?? = v ?? v ?? 2 + ?? + 1 and t a n ? ?? = ( ?? - 3 + ?? - 2 + ?? - 1 ) 1 2 , 0 < ?? , ?? , ?? < ?? 2 , then A + B is equal to : (1) C (2) ?? - ?? (3) 2 ?? - ?? (4) ?? 2 - ?? Q2 - 2024 (30 Jan Shift 2) For ?? , ?? ? ( 0 , ?? 2 ), let 3 s i n ? ( ?? + ?? ) = 2 s i n ? ( ?? - ?? ) and a real number k be such that t a n ? ?? = kt a n ? ?? . Then the value of ?? is equal to : (1) - 2 3 (2) -5 (3) 2 3 (4) 5 Q3 - 2024 (01 Feb Shift 2) The number of solutions of the equation 4 sin 2 ? ?? - 4 c o s 3 ? ?? + 9 - 4 c o s ? ?? = 0 ; ?? ? [ - 2 ?? , 2 ?? ] is : (1) 1 (2) 3 (3) 2 (4) 0 Q4 - 2024 (27 Jan Shift 1) Let the set of all ?? ? ?? such that the equation c o s ? 2 ?? + ?? sin ? ?? = 2 ?? - 7 has a solution be [ ?? , ?? ] and ?? = t a n ? 9 ° - t a n ? 27 ° - 1 c o t ? 63 ° + t a n ? 81 ° , then pqr is equal to Q5 - 2024 (27 Jan Shift 2) If 2 t a n 2 ? ?? - 5 s e c ? ?? = 1 has exactly 7 solutions in the interval [ 0 , ???? 2 ], for the least value of ?? ? ?? then ? k = 1 n ? k 2 k is equal to : (1) 1 2 15 ( 2 14 - 14 ) (2) 1 2 14 ( 2 15 - 15 ) (3) 1 - 15 2 13 (4) 1 2 13 ( 2 14 - 15 ) Q6 - 2024 (29 Jan Shift 1) If ?? , - ?? 2 < ?? < ?? 2 is the solution of 4 c o s ? ?? + 5 s i n ? ?? = 1, then the value of t a n ? ?? is (1) 10 - v 10 6 (2) 10 - v 10 12 (3) v 10 - 10 12 (4) v 10 - 10 6 Q7- 2024 (29 Jan Shift 2) The sum of the solutions ?? ? R of the equation 3 c o s ? 2 ?? + c o s 3 ? 2 ?? c o s 6 ? ?? - s i n 6 ? ?? = ?? 3 - ?? 2 + 6 is (1) 0 (2) 1 (3) -1 (4) 3 Q8 - 2024 (30 Jan Shift 1) If 2 sin 3 ? ?? + sin ? 2 ?? c o s ? ?? + 4 s i n ? ?? - 4 = 0 has exactly 3 solutions in the interval [ 0 , n ?? 2 ] , n ? N, then the roots of the equation ?? 2 + ???? + ( ?? - 3 ) = 0 belong to : (1) ( 0 , 8 ) (2) ( - 8 , 0 ) (3) ( - v 17 2 , v 17 2 ) (4) Z Answer Key Q1 (1) Q2 (2) Q3 (4) Q4 (48) Q5 (4) Q6 (3) Q7 (3) Q8 (2) Solutions Q1 Finding t a n ? ( ?? + ?? ) we get ? ? t a n ? ( A + B ) = t a n ? ?? + t a n ? ?? 1 - t a n ? ?? t a n ? ?? = 1 v ?? ( ?? 2 + ?? + 1 ) + v ?? v ?? 2 + ?? + 1 1 - 1 ?? 2 + ?? + 1 ? ? t a n ? ( A + B ) = ( 1 + ?? ) ( v ?? 2 + ?? + 1 ) ( ?? 2 + ?? ) ( v ?? ) ( 1 + ?? ) ( v ?? 2 + ?? + 1 ) ( ?? 2 + ?? ) ( v ?? ) t a n ? ( ?? + ?? ) = v ?? 2 + ?? + 1 ?? v ?? = t a n ? ?? ?? + ?? = ?? Q2 3 s i n ? ?? c o s ? ?? + 3 s i n ? ?? c o s ? ?? = 2 s i n ? ?? c o s ? ?? - 2 s i n ? ?? c o s ? ?? 5 s i n ? ?? c o s ? ?? = - sin ? ?? c o s ? ?? t a n ? ?? = - 1 5 t a n ? ?? t a n ? ?? = - 5 t a n ? ?? Q3 4 sin 2 ? ?? - 4 c o s 3 ? ?? + 9 - 4 c o s ? ?? = 0 ; ?? ? [ - 2 ?? , 2 ?? ] 4 - 4 c o s 2 ? ?? - 4 c o s 3 ? ?? + 9 - 4 c o s ? ?? = 0 4 c o s 3 ? ?? + 4 c o s 2 ? ?? + 4 c o s ? ?? - 13 = 0 4 c o s 3 ? ?? + 4 c o s 2 ? ?? + 4 c o s ? ?? = 13 L.H.S. = 12 c a n ' t be equal to 13 . Q4 c o s ? 2 ?? + ?? · sin ? ?? = 2 ?? - 7 ?? ( sin ? ?? - 2 ) = 2 ( sin ? ?? - 2 ) ( sin ? ?? + 2 ) sin ? ?? = 2 , ? ?? = 2 ( sin ? ?? + 2 ) ? ?? ? [ 2 , 6 ] ?? = 2 ? ?? = 6 ?? = t a n ? 9 ° + c o t ? 9 ° - t a n ? 27 - c o t ? 27 ?? = 1 sin ? 9 · c o s ? 9 - 1 sin ? 27 · c o s ? 27 = 2 [ 4 v 5 - 1 - 4 v 5 + 1 ] ?? = 4 p.q. . ?? = 2 × 6 × 4 = 48 Q5 2 t a n 2 ? ?? - 5 s e c ? ?? - 1 = 0 ? ? 2 s e c 2 ? ?? - 5 s e c ? ?? - 3 = 0 ? ? ( 2 s e c ? ?? + 1 ) ( s e c ? ?? - 3 ) = 0 ? ? s e c ? ?? = - 1 2 , 3 ? ? c o s ? ?? = - 2 , 1 3 ? ? c o s ? ?? = 1 3 For 7 solutions n = 13 So, ? k = 1 13 ? k 2 k = S (say) S = 1 2 + 2 2 2 + 3 2 3 + ? + 13 2 13 1 2 ?? = 1 2 2 + 1 2 3 + ? . + 12 2 13 + 13 2 14 ? ?? 2 = 1 2 · 1 - 1 2 13 1 - 1 2 - 13 2 14 ? ?? = 2 · ( 2 13 - 1 2 13 ) - 13 2 13 Q6 4 + 5 t a n ? ?? = s e c ? ?? Squaring : 24 t a n 2 ? ?? + 4 0 t a n ? ?? + 15 = 0 t a n ? ?? = - 10 ± v 10 12 and t a n ? ?? = - ( 10 + v 10 12 ) is Rejected.Read More
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FAQs on Trigonometric Ratios, Functions & Equations: JEE Mains Previous Year Questions (2021-2024) - Mathematics (Maths) for JEE Main & Advanced
| 1. How do you find the trigonometric ratios of an angle in a right triangle? |  |
Ans. To find the trigonometric ratios of an angle in a right triangle, you can use the ratios of the sides of the triangle. The sine ratio is the ratio of the length of the side opposite the angle to the hypotenuse, the cosine ratio is the ratio of the length of the side adjacent to the angle to the hypotenuse, and the tangent ratio is the ratio of the length of the side opposite the angle to the side adjacent to the angle.
| 2. What are the properties of trigonometric functions? | |
Ans. Trigonometric functions have various properties, including periodicity, symmetry, and the relationships between different functions like sine, cosine, and tangent. They also have specific ranges and domains, such as sine and cosine functions having a range of [-1,1] and tangent having a domain excluding odd multiples of pi/2.
| 3. How do you solve trigonometric equations? | |
Ans. To solve trigonometric equations, you can use algebraic techniques along with trigonometric identities and properties. You may need to simplify the equation, use trigonometric identities to rewrite it in a more manageable form, and then solve for the variable by isolating it on one side of the equation.
| 4. What are the common trigonometric identities used in trigonometry? | |
Ans. Some common trigonometric identities used in trigonometry include Pythagorean identities (sin^2θ + cos^2θ = 1), reciprocal identities (cscθ = 1/sinθ, secθ = 1/cosθ, cotθ = 1/tanθ), and double angle identities (sin2θ = 2sinθcosθ).
| 5. How are trigonometric functions used in real-life applications? | |
Ans. Trigonometric functions are used in various real-life applications such as engineering, physics, and astronomy. They help in calculating distances, angles, and heights in structures, analyzing wave patterns, and predicting celestial events.
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