Trigonometric Equations — JEE Main Mathematics MCQs with Solutions
Free JEE Main Mathematics Trigonometric Equations MCQs with step-by-step solutions (6 questions). Part of Trigonometric Ratios. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — 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.
Q2 — 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.
Q3 — 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.
Q4 — Trigonometric Equations · medium
$x \tan^2 120° + 4 \cos^2 150° = 9$ ⇒ x =
A. 3 ✓ Correct
B. 1
C. 2
D. 4
Solution: Substitute: $\tan 120° = -\sqrt{3}$, $\cos 150° = -\frac{\sqrt{3}}{2}$.
Q5 — Trigonometric Equations · hard
$1 + \cos x + \cos^2 x + ....... $ to $∞ = 4 + 2\sqrt{3}$ ⇒ x =
A. 30°
B. 60° ✓ Correct
C. 45°
D. 90°
Solution: Sum of GP: $\frac{1}{1 - \cos x} = 4 + 2\sqrt{3}$.
Q6 — Trigonometric Equations · hard
$1 + \sin x + \sin^2 x + ........ $ to $∞ = 4 + 2\sqrt{3}$ ⇒ x =
A. 30°, 60°
B. 60°, 120° ✓ Correct
C. 90°, 120°
D. 30°, 45°
Solution: Solve $\frac{1}{1 - \sin x} = 4 + 2\sqrt{3}$.