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JEE • MATHEMATICS • TRIGONOMETRY RATIOS

Trigonometry ratios practice · Medium

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 1-30 of 157 questions
1 Medium
1 of 157
Analyzing +4 | -1
If tan⁡α=pq\tan \alpha=\frac{p}{q}, where α=6β\alpha=6 \beta and α\alpha being an acute angle, then 12[pcosec⁡2β−qsec⁡2β]\frac{1}{2}[p \operatorname{cosec} 2 \beta-q \sec 2 \beta] is equal to
View answer and explanation
Correct option: A
p2+q2\sqrt{p^2+q^2}
Option A (p2+q2\sqrt{{p^2 + q^2}}) is correct because the expression simplifies to this value using trigonometric identities and the given relation tan⁡α=pq\tan \alpha = \frac{{p}}{{q}}.
2 Medium
2 of 157
Analyzing +4 | -1
If the △ABC\triangle A B C is acute angled at CC, then
View answer and explanation
Correct option: A
cos⁡2A+cos⁡2B−cos⁡2C<1\cos 2 A+\cos 2 B-\cos 2 C<1
Option A (cos⁡2A+cos⁡2B−cos⁡2C<1\cos 2 A+\cos 2 B-\cos 2 C<1) is correct because, in an acute angled triangle, the given inequality holds true based on trigonometric identities.
3 Medium
3 of 157
Application +4 | -1
sin⁡2A2+sin⁡2B2+sin⁡2C2=\sin ^2 \frac{A}{2}+\sin ^2 \frac{B}{2}+\sin ^2 \frac{C}{2}=
View answer and explanation
Correct option: B
1−2sin⁡A2sin⁡B2sin⁡C21-2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}
Option B (1−2sin⁡A2sin⁡B2sin⁡C21-2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}) is correct because by applying half-angle formulas and trigonometric identities, the expression simplifies to the given form.
4 Medium
4 of 157
Application +4 | -1
sin⁡2(A/2)+sin⁡2(B/2)−sin⁡2(C/2)=\sin ^2(A / 2)+\sin ^2(B / 2)-\sin ^2(C / 2)=
View answer and explanation
Correct option: A
1−2cos⁡A2cos⁡B2sin⁡C21-2 \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}
Option A (1−2cos⁡A2cos⁡B2sin⁡C21-2 \cos \frac{{A}}{{2}} \cos \frac{{B}}{{2}} \sin \frac{{C}}{{2}}) is correct because it is the result of applying trigonometric identities to simplify the given expression.
5 Medium
5 of 157
Application +4 | -1
If A+B+C=180∘(A,B>0)A+B+C=180^{\circ}(A, B > 0) and the angle CC is obtuse, then
View answer and explanation
Correct option: B
tan⁡Atan⁡B<1\tan A \tan B < 1
Option B (tan⁡Atan⁡B<1\tan A \tan B < 1) is correct because when C is obtuse, tan⁡C\tan C is negative, leading to the inequality tan⁡Atan⁡B<1\tan A \tan B < 1.
6 Medium
6 of 157
Application +4 | -1
In a △ABC\triangle A B C, if a=8,b=15,c=17a=8, b=15, c=17, then sin⁡A2\sin \frac{A}{2} and cos⁡A\cos A are equal to
View answer and explanation
Correct option: A
117,1517\frac{1}{\sqrt{17}}, \frac{15}{17}
Option A (117,1517\frac{1}{\sqrt{17}}, \frac{15}{17}) is correct because by using the given side lengths, we can calculate the semi-perimeter and then apply the half-angle formula for sin⁡A2\sin \frac{A}{2} and the double-angle formula for cos⁡A\cos A.
7 Medium
7 of 157
Application +4 | -1
The value of cos⁡2π16+cos⁡23π16+cos⁡25π16+cos⁡27π16\cos ^2 \frac{\pi}{16}+\cos ^2 \frac{3 \pi}{16}+\cos ^2 \frac{5 \pi}{16}+\cos ^2 \frac{7 \pi}{16} is
View answer and explanation
Correct option: A
2
Option A (2) is correct because by using the identity cos⁡(π2−x)=sin⁡x\cos(\frac{\pi}{2} - x) = \sin x, we can pair the terms and simplify using sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1.
8 Medium
8 of 157
Application +4 | -1
If AA lies in the second quadrant and 3tan⁡A+4=03 \tan A+4=0, the value of 2cot⁡A−5cos⁡A+sin⁡A2 \cot A-5 \cos A+\sin A is
View answer and explanation
Correct option: D
2310\frac{23}{10}
Option D (2310\frac{{23}}{{10}}) is correct because substituting the values of sin A, cos A, and cot A, derived from the given tangent value and quadrant information, into the expression yields 2310\frac{{23}}{{10}}.
9 Medium
9 of 157
Application +4 | -1
A quadratic equation whose roots are cosec⁡2θ\operatorname{cosec}^2 \theta and sec⁡2θ\sec ^2 \theta, can be
View answer and explanation
Correct option: C
x2−5x+5=0x^2-5 x+5=0
Option C (x2−5x+5=0x^2-5 x+5=0) is correct because the roots cosec⁡2θ\operatorname{cosec}^2 \theta and sec⁡2θ\sec ^2 \theta satisfy this equation.
10 Medium
10 of 157
Application +4 | -1
In a △ABC,∠C=60∘\triangle A B C, \angle C=60^{\circ}, then 1a+c+1b+c\frac{1}{a+c}+\frac{1}{b+c} is equal to
View answer and explanation
Correct option: C
3a+b+c\frac{3}{a+b+c}
Option C (3a+b+c\frac{3}{a+b+c}) is correct. By using the cosine rule with ∠C=60∘\angle C = 60^{\circ} and manipulating the resulting equation, the given expression can be simplified to 3a+b+c\frac{3}{a+b+c}.
11 Medium
11 of 157
Application +4 | -1
If tan⁡α=a\tan \alpha=\sqrt{ a}, where aa is a rational number which is not a perfect square, then which of the following is a rational number?
View answer and explanation
Correct option: C
cos⁡2α\cos 2 \alpha
Option C (cos⁡2α\cos 2 \alpha) is correct because cos⁡2α\cos 2 \alpha can be expressed as 1−tan⁡2α1+tan⁡2α\frac{{1-\tan^2 \alpha}}{{1+\tan^2 \alpha}}, which results in a rational number when tan⁡α=a\tan \alpha = \sqrt{{a}} and aa is rational.
12 Medium
12 of 157
Application +4 | -1
If θ=π2n+1\theta=\frac{\pi}{2^n+1} then the value of cos⁡θcos⁡2θcos⁡22θ……cos⁡2n−1θ\cos \theta \cos 2 \theta \cos 2^2 \theta \ldots \ldots \cos 2^{n-1} \theta is equal to -
View answer and explanation
Correct option: B
1/2n1 / 2^n
Option B (1/2n1 / 2^n) is correct because the expression simplifies to sin⁡2nθ2nsin⁡θ\frac{{\sin 2^n \theta}}{{2^n \sin \theta}}, which then simplifies to 1/2n1 / 2^n using the given condition.
13 Medium
13 of 157
Application +4 | -1
If cos⁡A3=cos⁡B4=15,   −π2<A<0,   −π2<B<0\frac{{\cos A}}{3} = \frac{{\cos B}}{4} = \frac{1}{5},\;\, - \frac{\pi }{2} < A < 0,\,\; - \frac{\pi }{2} < B < 0 then the value of 2sin⁡A+4sin⁡B2\sin A + 4\sin B is

View answer and explanation
Correct option: C
−4- 4
Option C (-4) is correct because after finding the values of sin A and sin B in the fourth quadrant, substituting them into the expression gives -4.
14 Medium
14 of 157
Application +4 | -1
The value of sin⁡  10∘+sin⁡  20∘+sin⁡  30∘+...+sin⁡  360∘\sin \,\,10^\circ + \sin \,\,20^\circ + \sin \,\,30^\circ + ... + \sin \,\,360^\circ is equal to
View answer and explanation
Correct option: C
3\sqrt 3
Option A (0\mathrm{0}) is correct because the sum of sines of angles in arithmetic progression over a full cycle is zero.
15 Medium
15 of 157
Application +4 | -1
The value of cos⁡2(3π5)+cos⁡2(4π5)\cos ^2\left(\frac{3 \pi}{5}\right)+\cos ^2\left(\frac{4 \pi}{5}\right) is equal to
View answer and explanation
Correct option: A
3/43 / 4
Option A (3/4) is correct because by converting the angles to degrees and using trigonometric identities, the expression simplifies to 3/4.

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16 Medium
16 of 157
Analyzing +4 | -1
If   A=sin⁡2θ+cos⁡4θ\;A = {\sin ^2}\mathrm{\theta } + {\cos ^4}\mathrm{\theta } then for all real values of θ\mathrm{\theta }

View answer and explanation
Correct option: B
34≤A≤1\frac{3}{4} \le A \le 1
Option B (34≤A≤1\frac{3}{4} \le A \le 1) is correct because the minimum value of AA is 34\frac{3}{4} and the maximum value is 1, which can be found by completing the square and using trigonometric properties.
17 Medium
17 of 157
Application +4 | -1
If tan⁡α=1/7\tan \alpha=1 / 7 and sin⁡β=1/10\sin \beta=1 / \sqrt{10} where 0<α,β<π/20 < \alpha, \beta < \pi / 2, then 2β2 \beta is equal to
View answer and explanation
Correct option: A
π/4−α\pi / 4-\alpha
Option A (π/4−α\pi / 4 - \alpha) is correct because it represents the value of 2β2\beta given the conditions tan⁡α=1/7\tan \alpha = 1/7 and sin⁡β=1/10\sin \beta = 1/\sqrt{{10}}, derived using trigonometric identities.
18 Medium
18 of 157
Application +4 | -1
Interior angles of the regular dodecagon is
View answer and explanation
Correct option: D
5π/65 \pi / 6
Option D (5π/65 \pi / 6) is correct because the interior angle of a regular dodecagon (12 sides) is calculated using the formula (n−2)×180∘/n(n-2) \times 180^{\circ} / n.
19 Medium
19 of 157
Application +4 | -1
(sin⁡θ+cosec⁡θ)2+(cos⁡θ+sec⁡θ)2(\sin \theta+\operatorname{cosec} \theta)^2+(\cos \theta+\sec \theta)^2 is
View answer and explanation
Correct option: A
≥9\geq 9
Option A (≥9\geq 9) is correct because the given expression can be simplified and shown to be greater than or equal to 9 using AM-GM inequality.
20 Medium
20 of 157
Application +4 | -1
The maximum value of acos⁡x+bsin⁡xa \cos x+b \sin x is
View answer and explanation
Correct option: D
(a2+b2)1/2\left(a^2+b^2\right)^{1 / 2}
Option D ((a2+b2)1/2\left(a^2+b^2\right)^{1 / 2}) is correct because the maximum value of acos⁡x+bsin⁡xa \cos x + b \sin x is a2+b2\sqrt{{a^2 + b^2}}.
21 Medium
21 of 157
Application +4 | -1
The resultant of three forces represented in magnitude and direction by the sides of a triangle ABC{ABC} taken in order with BC=5 cm,CA=5 cm{BC}=5 {~cm}, {CA}=5 {~cm} and AB=8 cm{AB}=8 {~cm}, is a couple of moment
View answer and explanation
Correct option: B
24 units
Option B (24 units) is correct because the resultant of the forces is a couple with a moment equal to twice the area of the triangle.
22 Medium
22 of 157
Application +4 | -1
In a triangle tan⁡A+tan⁡B+tan⁡C=6\tan A+\tan B+\tan C=6 and tan⁡A\tan A tan⁡B=2\tan B=2, then the values of tan⁡A,tan⁡B\tan A, \tan B and tan⁡C\tan C are
View answer and explanation
Correct option: B
2,1,32,1,3
Option B (2,1,3) is correct because it satisfies the given conditions: tan⁡A+tan⁡B+tan⁡C=6\tan A + \tan B + \tan C = 6 and tan⁡Atan⁡B=2\tan A \tan B = 2.
23 Medium
23 of 157
Application +4 | -1
If sin⁡2θ+sin⁡2ϕ=1/2\sin 2 \theta+\sin 2 \phi=1 / 2 and cos⁡2θ+cos⁡2ϕ=3/2\cos 2 \theta+\cos 2 \phi=3 / 2, then cos⁡2(θ−ϕ)=\cos ^2(\theta-\phi)=
View answer and explanation
Correct option: B
5/85 / 8
Option B (5 / 8) is correct because squaring and adding the given equations, then using trigonometric identities, leads to cos⁡2(θ−ϕ)=5/8\cos^2(\theta - \phi) = 5/8.
24 Medium
24 of 157
Application +4 | -1
If ∑i=1nsin⁡θi=n\sum\limits_{{i=1}}^n \sin \theta_i=n, then the value of ∑i=1ncos⁡θi\sum\limits_{{i=1}}^n \cos \theta_i is
View answer and explanation
Correct option: B
0
Option B (0) is correct because if ∑i=1nsin⁡θi=n\sum_{i=1}^n \sin \theta_i = n, then each sin⁡θi\sin \theta_i must be 1, which means each θi=π2\theta_i = \frac{\pi}{2}. Therefore, each cos⁡θi=0\cos \theta_i = 0, and ∑i=1ncos⁡θi=0\sum_{i=1}^n \cos \theta_i = 0.
25 Medium
25 of 157
Analyzing +4 | -1
Let f(x)=1+2cos⁡x+3sin⁡xf(x)=1+2 \cos x+3 \sin x. If αf(x)+βf(x−γ)=1\alpha f(x)+\beta f(x-\gamma)=1 where α,β,γ∈R\alpha, \beta, \gamma \in R holds for all x∈Rx \in R, then βcos⁡γα\frac{\beta \cos \gamma}{\alpha} is equal to
View answer and explanation
Correct option: A
-1
Option A (-1) is correct because by finding f(x-π) and comparing αf(x) + βf(x-γ) = 1, we get α = β = 1/2 and γ = π, thus βcosγ/α = -1.
26 Medium
26 of 157
Application +4 | -1
If x=acos⁡2θsin⁡θx=a \cos ^2 \theta \sin \theta and y=y= a sin⁡2θcos⁡θ\sin ^2 \theta \cos \theta, then (x2+y2)3\left(x^2+y^2\right)^3 is
View answer and explanation
Correct option: B
a2x2y2a^2 x^2 y^2
Option B (a2x2y2a^2 x^2 y^2) is correct because by substituting the given expressions for x and y, and simplifying, we find that (x2+y2)3=a2x2y2(x^2+y^2)^3 = a^2 x^2 y^2.
27 Medium
27 of 157
Application +4 | -1
If θ\theta and ϕ\phi are acute angles such that sin⁡θ=1/2\sin \theta=1 / 2 and cos⁡ϕ=1/3\cos \phi=1 / 3, then θ+ϕ\theta+\phi lies is-
View answer and explanation
Correct option: B
π/2,2π/3\pi / 2,2 \pi / 3
Option B (π/2,2π/3\pi / 2,2 \pi / 3) is correct because given the acute angles and their respective sine and cosine values, the sum of the angles lies within this interval.
28 Medium
28 of 157
Application +4 | -1
sin⁡36∘sin⁡72∘sin⁡108∘sin⁡144∘=\sin 36^{\circ} \sin 72^{\circ} \sin 108^{\circ} \sin 144^{\circ}=
View answer and explanation
Correct option: D
5/165 / 16
Option D (5/165 / 16) is correct because after applying trigonometric identities and using known values, the expression simplifies to 5/165/16.
29 Medium
29 of 157
Application +4 | -1
In △ABC\triangle A B C, if sin⁡2A+sin⁡2B=sin⁡2C\sin ^{2} A+\sin ^{2} B=\sin ^{2} C, then ∠C=\angle C=
View answer and explanation
Correct option: D
90∘90^{\circ}
Option D (90°) is correct because the given equation sin2(A) + sin2(B) = sin2(C) implies a2 + b2 = c2, which is the Pythagorean theorem, indicating that angle C is 90 degrees.
30 Medium
30 of 157
Application +4 | -1
If Tn=cos⁡n  θ+sin⁡n  θ{T_n} = {\cos ^n}\,\,\theta + {\sin ^n}\,\,\theta, then 2 T6−3 T4+1=2\,{T_6} - 3\,{T_4} + 1 =
View answer and explanation
Correct option: C
0\mathrm{0}
Option C (0) is correct because the expression 2T_6 - 3T_4 + 1 simplifies to 0 using trigonometric identities and the given definition of T_n.
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