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Trigonometric Ratios, Functions & Equations: JEE Main Previous Year Questions (2021-2026)

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JEE Main Previous Year Questions (2021-2026): 
Trigonometric Ratios, Functions & Equations  
 
(January 2026) 
 
 
Trigonometric Ratio and Identities 
 
Q1:  
A: tan A, tan C, tan B are in A.P. 
B: tan A, tan C, tan B are in G.P. 
C: tan A, tan B, tan C are in G.P. 
D: tan A, tan B, tan C are in A.P. 
Answer: B 
Explanation: 
Given: 
 
Substitute back into (1) 
Page 2


JEE Main Previous Year Questions (2021-2026): 
Trigonometric Ratios, Functions & Equations  
 
(January 2026) 
 
 
Trigonometric Ratio and Identities 
 
Q1:  
A: tan A, tan C, tan B are in A.P. 
B: tan A, tan C, tan B are in G.P. 
C: tan A, tan B, tan C are in G.P. 
D: tan A, tan B, tan C are in A.P. 
Answer: B 
Explanation: 
Given: 
 
Substitute back into (1) 
 
? tan²C + tan²C tan A tan B = tan A tan B + tan²C tan A tan B 
? tan²C = tan A tan B 
? tan A, tan C, tan B are in G.P. 
 
Q2: The value of  is equal to 
A: 32 
B: 64 
C: 12 
D: 16 
Answer: B 
Explanation: 
Page 3


JEE Main Previous Year Questions (2021-2026): 
Trigonometric Ratios, Functions & Equations  
 
(January 2026) 
 
 
Trigonometric Ratio and Identities 
 
Q1:  
A: tan A, tan C, tan B are in A.P. 
B: tan A, tan C, tan B are in G.P. 
C: tan A, tan B, tan C are in G.P. 
D: tan A, tan B, tan C are in A.P. 
Answer: B 
Explanation: 
Given: 
 
Substitute back into (1) 
 
? tan²C + tan²C tan A tan B = tan A tan B + tan²C tan A tan B 
? tan²C = tan A tan B 
? tan A, tan C, tan B are in G.P. 
 
Q2: The value of  is equal to 
A: 32 
B: 64 
C: 12 
D: 16 
Answer: B 
Explanation: 
 
 
Q3: If cot x = 5/12 for some x ? (p, 3p/2), then sin 7x (cos 13x/2 + sin 13x/2) + cos 7x (cos 
13x/2 - sin 13x/2) is equal to: 
A: 5/v13 
B: 6/v26 
C: 4/v26 
D: 1/v13 
Answer: D 
Explanation: 
 
Rearranging terms and use sin A cos B - cos A sin B = sin(A - B) and cos A cos B + sin A sin B 
Page 4


JEE Main Previous Year Questions (2021-2026): 
Trigonometric Ratios, Functions & Equations  
 
(January 2026) 
 
 
Trigonometric Ratio and Identities 
 
Q1:  
A: tan A, tan C, tan B are in A.P. 
B: tan A, tan C, tan B are in G.P. 
C: tan A, tan B, tan C are in G.P. 
D: tan A, tan B, tan C are in A.P. 
Answer: B 
Explanation: 
Given: 
 
Substitute back into (1) 
 
? tan²C + tan²C tan A tan B = tan A tan B + tan²C tan A tan B 
? tan²C = tan A tan B 
? tan A, tan C, tan B are in G.P. 
 
Q2: The value of  is equal to 
A: 32 
B: 64 
C: 12 
D: 16 
Answer: B 
Explanation: 
 
 
Q3: If cot x = 5/12 for some x ? (p, 3p/2), then sin 7x (cos 13x/2 + sin 13x/2) + cos 7x (cos 
13x/2 - sin 13x/2) is equal to: 
A: 5/v13 
B: 6/v26 
C: 4/v26 
D: 1/v13 
Answer: D 
Explanation: 
 
Rearranging terms and use sin A cos B - cos A sin B = sin(A - B) and cos A cos B + sin A sin B 
= cos(A - B):
 
 
Q4: Let Then the value of 
sin(15?/2)(cos 8? + sin 8?) + cos(15?/2)(cos 8? - sin 8?) 
is equal to: 
A: (v2 - 1)/v3 
B: v2/v3 
C: (1 - v2)/v3 
D: -v2/v3 
Answer: C 
Page 5


JEE Main Previous Year Questions (2021-2026): 
Trigonometric Ratios, Functions & Equations  
 
(January 2026) 
 
 
Trigonometric Ratio and Identities 
 
Q1:  
A: tan A, tan C, tan B are in A.P. 
B: tan A, tan C, tan B are in G.P. 
C: tan A, tan B, tan C are in G.P. 
D: tan A, tan B, tan C are in A.P. 
Answer: B 
Explanation: 
Given: 
 
Substitute back into (1) 
 
? tan²C + tan²C tan A tan B = tan A tan B + tan²C tan A tan B 
? tan²C = tan A tan B 
? tan A, tan C, tan B are in G.P. 
 
Q2: The value of  is equal to 
A: 32 
B: 64 
C: 12 
D: 16 
Answer: B 
Explanation: 
 
 
Q3: If cot x = 5/12 for some x ? (p, 3p/2), then sin 7x (cos 13x/2 + sin 13x/2) + cos 7x (cos 
13x/2 - sin 13x/2) is equal to: 
A: 5/v13 
B: 6/v26 
C: 4/v26 
D: 1/v13 
Answer: D 
Explanation: 
 
Rearranging terms and use sin A cos B - cos A sin B = sin(A - B) and cos A cos B + sin A sin B 
= cos(A - B):
 
 
Q4: Let Then the value of 
sin(15?/2)(cos 8? + sin 8?) + cos(15?/2)(cos 8? - sin 8?) 
is equal to: 
A: (v2 - 1)/v3 
B: v2/v3 
C: (1 - v2)/v3 
D: -v2/v3 
Answer: C 
Explanation: 
 
 
 
Q5: The value of cosec 10° - v3 sec 10° is equal to: 
A: 2 
B: 6 
C: 8 
D: 4 
Answer: D 
Explanation:
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FAQs on Trigonometric Ratios, Functions & Equations: JEE Main Previous Year Questions (2021-2026)

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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