Page 1
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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