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Trigonometric Ratios — JEE Main Mathematics MCQs with Solutions
Free JEE Main Mathematics Trigonometric Ratios MCQs with step-by-step solutions covering Basic Identities and Operations, Trigonometric Equations, Special Angles and Values, Critical Reasoning, Ordering and Comparison. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — Basic Identities and Operations · easy
sec A + tan A = 3 ⇒ sec A =
A. $\frac{10}{3}$
B. $\frac{5}{3}$ ✓ Correct
C. $\frac{2}{3}$
D. $\frac{4}{3}$
Solution: Using $\sec^2 A - \tan^2 A = 1$ and the given relation.
Q2 — Basic Identities and Operations · easy
sec A – tan A = 4 ⇒ tan A =
A. $\frac{15}{8}$
B. $-\frac{15}{8}$ ✓ Correct
C. $\frac{17}{8}$
D. $-\frac{17}{8}$
Solution: Apply the identity $(\sec A + \tan A)(\sec A - \tan A) = 1$.
Q3 — Basic Identities and Operations · easy
sec A – tan A = 5 ⇒ sin A =
A. $\frac{6}{13}$
B. $-\frac{6}{13}$
C. $\frac{12}{13}$ ✓ Correct
D. $-\frac{12}{13}$
Solution: From $\sec A - \tan A = 5$, find $\sec A$ then $\sin A$.
Q4 — Basic Identities and Operations · easy
cosec A – cot A = 5 ⇒ cosec A =
A. $\frac{5}{13}$
B. $\frac{13}{5}$ ✓ Correct
C. $-\frac{5}{13}$
D. $-\frac{13}{5}$
Solution: Use $(\cosec A - \cot A)(\cosec A + \cot A) = 1$.
Q5 — Basic Identities and Operations · easy
cosec A – cot A = 6 ⇒ cot A =
A. $\frac{12}{35}$
B. $-\frac{12}{35}$ ✓ Correct
C. $\frac{35}{12}$
D. $-\frac{35}{12}$
Solution: Solve using $\cosec^2 A - \cot^2 A = 1$.
Q6 — Special Angles and Values · hard
tan 20° + tan 40° + tan 60° + ...... + tan 180° =
A. 0 ✓ Correct
B. 1
C. 2
D. 3
Solution: Use periodicity and complementary angle properties.
Q7 — Basic Identities and Operations · hard
$3[\sin x - \cos x]^4 + 6[\sin x + \cos x]^2 + 4[\sin^6 x + \cos^6 x]$ =
A. 3
B. 6
C. 4
D. 13 ✓ Correct
Solution: Expand and simplify using algebraic identities.
Q8 — Basic Identities and Operations · hard
$\frac{\cos^3 A - \sin^3 A}{\cos A - \sin A} + \frac{\cos^3 A + \sin^3 A}{\cos A + \sin A} = k$, then k =
A. 0
B. 1
C. 2 ✓ Correct
D. –1
Solution: Factor and simplify the algebraic expressions.
Q9 — Basic Identities and Operations · hard
$\frac{\sin^2 α + \tan^2 α \cos α}{1 + \cot^2 α(1 + \tan^2 α)}$ =
A. –1
B. 0
C. 1 ✓ Correct
D. 2
Solution: Simplify using Pythagorean identities.
Q10 — Basic Identities and Operations · hard
$\frac{1}{(1 + \cot^2 α)^2} + \frac{\tan^2 α}{(1 + \tan^2 α)^2} + \frac{1}{1 + \tan^2 α}$ =
A. –1
B. 0
C. 1 ✓ Correct
D. 2
Solution: Use $1 + \cot^2 α = \cosec^2 α$ and $1 + \tan^2 α = \sec^2 α$.
Q11 — Basic Identities and Operations · hard
$\frac{(\sqrt{3} + 2\cos α)^3 + (1 + 2\sin α)^3}{(1 - 2\sin α)^3} + \frac{(\sqrt{3} - 2\cos α)^3}{(\sqrt{3} - 2\sin α)^3}$ =
A. 1
B. $\sqrt{3}$
C. 0 ✓ Correct
D. –1
Solution: Expand and cancel terms carefully.
Q12 — Special Angles and Values · hard
cos(40° + θ) cos(120° + θ) + cos(220° + θ) + cos(300° + θ) =
A. 3
B. 2
C. 1
D. 0 ✓ Correct
Solution: Group terms using complementary angles: $(40° + θ) + (220° + θ) = 260°$, $(120° + θ) + (300° + θ) = 420°$.
Q13 — Trigonometric Equations · hard
If $\tan θ = \frac{\sin α - \cos α}{\sin α + \cos α}$, $\sin α - \cos α = k \sin θ$, then k =
A. 1
B. $\sqrt{3}$
C. $\sqrt{2}$ ✓ Correct
D. 1/2
Solution: From the given tan, derive the relationship.
Q14 — Trigonometric Equations · hard
If $\tan θ = \frac{\sin α - \cos α}{\sin α + \cos α}$, $\sin α + \cos α = k \cos θ$, then k =
A. –1
B. 1
C. $\sqrt{2}$ ✓ Correct
D. $\sqrt{3}$
Solution: Calculate k from the given condition.
Q15 — Basic Identities and Operations · hard
$\frac{\cot θ + \cosec θ - 1}{\cot θ - \cosec θ + 1}$ I. (1) $\frac{1 - \cos θ}{\sin θ}$ II. (2) $\frac{\sin θ}{1 - \cos θ}$
A. $\frac{1 - \cos θ}{\sin θ}$ and $\frac{\sin θ}{1 - \cos θ}$ ✓ Correct
B. Other
C. Other
D. Other
Solution: Rationalize and simplify the expression.
Q16 — Basic Identities and Operations · hard
$\frac{1 + \cot θ + \cosec θ}{1 - \cot θ + \cosec θ}$ I. (1) $\frac{\sin α}{1 + \cos α}$ II. (1) $\frac{1 + \cos α}{\sin α}$
A. Type I and II ✓ Correct
B. Other
C. Other
D. Other
Solution: Simplify each part separately.
Q17 — Basic Identities and Operations · hard
$\frac{\tan α + \sec α - 1}{\tan α - \sec α + 1}$ I. (1) $\frac{1 - \sin α}{\cos α}$ II. (1) $\frac{\cos α}{1 - \sin α}$ III. (1) $\sec α + \tan α$
A. All three ✓ Correct
B. Other
C. Other
D. Other
Solution: All three forms are equivalent.
Q18 — Basic Identities and Operations · hard
$\frac{1 + \tan α + \sec α}{1 - \tan α + \sec α}$ I. (1) $\frac{1 - \sin α}{\cos α}$ II. (1) $\frac{\cos α}{1 - \sin α}$
A. Type I and II ✓ Correct
B. Other
C. Other
D. Other
Solution: Work through each simplification.
Q19 — Special Angles and Values · hard
$\cos^2 \frac{π}{18} + \cos^2 \frac{π}{9} + \cos^2 \frac{7π}{18} + \cos^2 \frac{4π}{9}$ =
A. 1
B. 2 ✓ Correct
C. 3
D. 4
Solution: Use symmetry of angles: $\frac{π}{18} + \frac{7π}{18} = \frac{π}{2}$.
Q20 — Special Angles and Values · hard
$\frac{\sin^2 \frac{π}{18} + \sin^2 \frac{π}{9} + \sin^2 \frac{7π}{18} + \sin^2 \frac{4π}{9}}{\cos^2 \frac{π}{18} + \cos^2 \frac{π}{9} + \cos^2 \frac{7π}{18} + \cos^2 \frac{4π}{18}}$ =
A. 1 ✓ Correct
B. 2
C. 3
D. 4
Solution: Use $\sin^2 θ + \cos^2 θ = 1$ strategically.
Q21 — Special Angles and Values · hard
$1 + \cos \frac{π}{7} + \cos \frac{2π}{7} + \cos \frac{3π}{7} + \cos \frac{4π}{7} + \cos \frac{5π}{7} + \cos \frac{6π}{7}$ =
A. 0 ✓ Correct
B. 1
C. 2
D. 3
Solution: Use symmetry about $\frac{π}{2}$.
Q22 — Special Angles and Values · hard
$\tan \frac{π}{24} \cdot \tan \frac{3π}{24} \cdot \tan \frac{5π}{24} \cdot \tan \frac{7π}{24} \cdot \tan \frac{9π}{24} \cdot \tan \frac{11π}{24} \cdot \tan \frac{16π}{24}$ =
A. 1
B. –1
C. $\sqrt{3}$
D. $-\sqrt{3}$ ✓ Correct
Solution: Use pairing and periodicity of tan.
Q23 — Special Angles and Values · hard
$\cot \frac{π}{20} \cdot \cot \frac{3π}{20} \cdot \cot \frac{5π}{20} \cdot \cot \frac{7π}{20} \cdot \cot \frac{9π}{20} \cdot \cot \frac{15π}{20}$ =
A. 1
B. –1 ✓ Correct
C. $\sqrt{3}$
D. $-\sqrt{3}$
Solution: Use complementary angle relationships.
Q24 — Basic Identities and Operations · hard
$(\sin α + \cosec α)^2 + (\sec α + \cos α)^2 = k + \tan^2 α + \cot^2 α$, k =
A. 9
B. 7 ✓ Correct
C. 5
D. 3
Solution: Expand and collect like terms.
Q25 — Basic Identities and Operations · hard
$\tan^2 θ - \sin^2 θ - \tan^2 θ \sin^2 θ$ =
A. 1
B. 0 ✓ Correct
C. 2
D. –1
Solution: Factor: $\tan^2 θ(1 - \sin^2 θ) - \sin^2 θ = \tan^2 θ \cos^2 θ - \sin^2 θ = \sin^2 θ - \sin^2 θ$.
Q26 — Trigonometric Equations · hard
If θ is not in 4th quadrant, $\tan θ = -4/3$ ⇒ $5 \sin θ + 10 \cos θ + 9 \sec θ + 16 \cosec θ + 4 \cot θ$ =
A. –1
B. 2/5
C. 4/5
D. 0 ✓ Correct
Solution: Substitute tan and solve using Pythagorean identity.
Q27 — Special Angles and Values · hard
A = 495° ⇒ $\cos^2 A + \sec^2 A - \sin^2 A - 2 \tan A$ =
A. 0
B. 1 ✓ Correct
C. 4
D. –1
Solution: First reduce 495° to a standard angle.
Q28 — Special Angles and Values · hard
$\frac{\sin 150° - 5\cos 300° - 7 \tan 225°}{\tan 135° - 3\sin 210°}$ =
A. 10
B. –5
C. –2
D. –3/2 ✓ Correct
Solution: Use standard angle values.
Q29 — Special Angles and Values · hard
$\frac{\sin(-660°)\tan(1050°)\sec(-420°)}{\cos(225°)\cosec(315°)\cos(510°)}$ =
A. $\frac{\sqrt{3}}{4}$
B. $\frac{\sqrt{3}}{2}$
C. $\frac{2}{\sqrt{3}}$
D. $\frac{4}{\sqrt{3}}$ ✓ Correct
Solution: Reduce all angles and use standard values.
Q30 — Special Angles and Values · hard
$\sin 160° \cos 110° + \sin 250° \cos 340° + \tan 110° \tan 340°$ =
A. –1
B. 1
C. 3
D. 0 ✓ Correct
Solution: Use supplementary and reference angle reductions.