Special Angles and Values — JEE Main Mathematics MCQs with Solutions
Free JEE Main Mathematics Special Angles and Values MCQs with step-by-step solutions (15 questions). Part of Trigonometric Ratios. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — 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.
Q2 — 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: $(40° + θ) + (220° + θ) = 260°$, $(120° + θ) + (300° + θ) = 420°$.
Q3 — Special Angles and Values · medium
If α and β are complementary angles, $\sin^2(90° - α) + \sin^2(90° - β)$ =
A. 0
B. 1 ✓ Correct
C. –1
D. 2
Solution: Use complementary angle substitutions.
Q4 — Special Angles and Values · medium
$\sin^2(51° - x) + \sin^2(39° - x)$ =
A. –1
B. 0
C. 1 ✓ Correct
D. 2
Solution: $(51° - x) + (39° - x) = 90° - 2x$, use complementary properties.
Q5 — 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: $\frac{π}{18} + \frac{7π}{18} = \frac{π}{2}$.
Q6 — 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.
Q7 — 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 $\frac{π}{2}$.
Q8 — 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.
Q9 — 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.
Q10 — 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.
Q11 — 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.
Q12 — 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.
Q13 — 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.
Q14 — Special Angles and Values · medium
$\cos 225° + \sin 165°$ =
A. $\sqrt{2}$
B. 0 ✓ Correct
C. 1
D. $\frac{\sqrt{3}}{2}$
Solution: $\cos 225° = -\frac{1}{\sqrt{2}}$, $\sin 165° = \sin 15° = \frac{1}{\sqrt{2}}$.
Q15 — Special Angles and Values · hard
$f(x) = x^3 - 2x^2 + 3x - 5 \Rightarrow f(\sin \frac{5π}{2}) + f(\sin \frac{3π}{2})$ =
A. 10
B. –10
C. 14
D. –14 ✓ Correct
Solution: $\sin \frac{5π}{2} = 1$, $\sin \frac{3π}{2} = -1$.