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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 = 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 $\sin A$.
Q2 — 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 $\cosec^2 A - \cot^2 A = 1$.
Q3 — 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 $(\cosec A - \cot A)(\cosec A + \cot A) = 1$.
Q4 — 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.
Q5 — 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 $(\sec A + \tan A)(\sec A - \tan A) = 1$.
Q6 — 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.
Q7 — 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.
Q8 — Basic Identities and Operations · hard
$\frac{1}{\sec 5x - \frac{1}{20x}} \implies \sec α + \tan α$ =
A. $5x$
B. $\frac{1}{20x}$
C. $10x$ or $-\frac{1}{10x}$
D. $10x$ or $\frac{1}{10x}$ ✓ Correct
Solution: From $\sec 5x - \tan 5x = \frac{1}{20x}$.
Q9 — Basic Identities and Operations · hard
$\tan P - \frac{1}{4p} \implies \sec θ - \tan θ$ =
A. $2p$ or $\frac{1}{2p}$
B. $\frac{1}{2p}$ or $-2p$
C. $-\frac{1}{2p}$ or $2p$
D. $-\frac{1}{2p}$ or $-2p$ ✓ Correct
Solution: Use $\sec^2 θ - \tan^2 θ = 1$.
Q10 — 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.
Q11 — Critical Reasoning · hard
$0 < θ < \frac{π}{2}$ ⇒
A. $\tan θ < θ < \sin θ$
B. $\tan θ > θ > \sin θ$
C. $\tan θ > θ < \sin θ$ ✓ Correct
D. $\tan θ < θ > \sin θ$
Solution: Compare growth rates in the first quadrant.
Q12 — Critical Reasoning · hard
$(\sqrt{1 - \cos^2 200°}) \cosec 200°$ =
A. –1
B. 0
C. 1 ✓ Correct
D. 2
Solution: $\sin 200° < 0$, so $|\sin 200°| = -\sin 200°$.
Q13 — Critical Reasoning · hard
If $\sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A$, then A lies in the quadrants
A. I, II
B. II, III
C. I, III ✓ Correct
D. I, IV
Solution: Work through quadrant analysis.
Q14 — 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.
Q15 — Basic Identities and Operations · hard
$\frac{\sqrt{1 + \cos A}}{\sqrt{1 - \cos A}} = k(\cosec A + \cot A) \Rightarrow k$ =
A. –1
B. 1
C. ±1 ✓ Correct
D. 2
Solution: Solve using identities.
Q16 — Critical Reasoning · hard
If $\tan θ = -\frac{4}{3}$, then $\sin θ$ =
A. $\frac{4}{5}$ but not $-\frac{4}{5}$
B. $\frac{4}{5}$ (or) $-\frac{4}{5}$ ✓ Correct
C. $-\frac{4}{5}$ but not $\frac{4}{5}$
D. $\frac{3}{5}$
Solution: tan is negative in II and IV quadrants.
Q17 — Critical Reasoning · hard
cos x + $\cos^2 x$ + $\cos^3 x$ = 1, $a \sin^6 x + b \sin^4 x + c \sin^2 x + d$ = 0 ⇒ $a + b + c + d$ =
A. 0 ✓ Correct
B. 1
C. –1
D. 2
Solution: Sum of coefficients.
Q18 — 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.
Q19 — Critical Reasoning · hard
x = cos 1°, y = cos 1 ⇒
A. x = y
B. x > y
C. x < y ✓ Correct
D. 2x = y
Solution: Since 1 rad = 57.3° and cos is decreasing.
Q20 — Critical Reasoning · hard
If $\sqrt{\frac{1 - \sin A}{1 + \sin A}} = \sec A - \tan A$, then A lies in the quadrants
A. I, II
B. II, III
C. I, IV ✓ Correct
D. I, III
Solution: Analyze sign conditions for the equation.
Q21 — Critical Reasoning · hard
If $k = (\sec A + \tan A)(\sec B + \tan B)(\sec C + \tan C) = (\sec A - \tan A)(\sec B - \tan B)(\sec C - \tan C)$, then k =
A. 0
B. ±1 ✓ Correct
C. ±3
D. ±4
Solution: Each product must equal 1.
Q22 — Critical Reasoning · hard
$\sin^6 \frac{49π}{49} + \cos^6 \frac{49π}{49} - 1 + 3\sin^2 \frac{49π}{49} \cos^2 \frac{49π}{49}$ =
A. 1
B. –1
C. 2
D. 0 ✓ Correct
Solution: Use $a^3 + b^3 = (a+b)^3 - 3ab(a+b)$ identity.
Q23 — Critical Reasoning · hard
$\frac{1}{2} \sin θ - \sqrt{\cot^2 θ - \cos^2 θ}$ =
A. $\cosec θ$
B. $\sec θ$
C. $\sin θ$ ✓ Correct
D. $\cos θ$
Solution: Simplify the radical first.
Q24 — 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.
Q25 — 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.
Q26 — 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: $\frac{π}{18} + \frac{7π}{18} = \frac{π}{2}$.
Q27 — 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.
Q28 — Critical Reasoning · hard
$\frac{\sqrt{1 - \cos θ} - \sqrt{1 + \cos θ}}{\sqrt{1 - \cos θ} + \sqrt{1 + \cos θ}}$ =
A. $-2 \cosec θ$ ✓ Correct
B. $2 \cosec θ$
C. $-2 \cot θ$
D. $2 \cot θ$
Solution: Manipulate using complementary properties.
Q29 — Critical Reasoning · hard
If x = a $\cos^2 θ$ sin θ and y = a $\sin^2 θ$ cos θ, then $\frac{(x^2 + y^2)^3}{x^2 y^2}$ =
A. a
B. $a^3$
C. $a^2$ ✓ Correct
D. $a^5$
Solution: Factor and simplify the ratio.
Q30 — 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.