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JEE • MATHEMATICS • TRIGONOMETRY RATIOS

Trigonometry ratios practice · Mixed difficulty

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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 394 questions
1 Medium
1 of 394
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 Easy
2 of 394
Application +4 | -1
The value of cos⁡12∘+cos⁡84∘+cos⁡156∘+cos⁡132∘\cos 12^{\circ}+\cos 84^{\circ}+\cos 156^{\circ}+\cos 132^{\circ} is
View answer and explanation
Correct option: B
−1/2-1 / 2
Option B (-1/2) is correct because by using the sum-to-product trigonometric identities, the expression simplifies to -1/2.
3 Medium
3 of 394
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.
4 Easy
4 of 394
Application +4 | -1
The value of the expression sin⁡6  θ+cos⁡6  θ+3sin⁡2  θ  cos⁡2  θ{\sin ^6}\,\,\theta + {\cos ^6}\,\,\theta + 3{\sin ^2}\,\,\theta \,\,{\cos ^2}\,\,\theta equals
View answer and explanation
Correct option: D
1
Option D (1) is correct because the expression is equivalent to (sin⁡2θ+cos⁡2θ)3(\sin^2 \theta + \cos^2 \theta)^3 which simplifies to 1.
5 Medium
5 of 394
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.
6 Medium
6 of 394
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.
7 Easy
7 of 394
Application +4 | -1
sin⁡2A1+cos⁡2A⋅cos⁡A1+cos⁡A=\frac{\sin 2 A}{1+\cos 2 A} \cdot \frac{\cos A}{1+\cos A}=
View answer and explanation
Correct option: A
tan⁡A2\tan \frac{A}{2}
Option A (tan⁡A2\tan \frac{{A}}{{2}}) is correct because the expression simplifies to tan⁡(A2)\tan(\frac{{A}}{{2}}) using trigonometric identities.
8 Medium
8 of 394
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.
9 Easy
9 of 394
Application +4 | -1
If sin⁡  x+cosec    x=2\sin \,\,x + \mathrm{cosec}\;\,\,x = 2 then, sin⁡n  x+cosecn   x{\sin ^n}\,\,x + \mathrm{cose}{\mathrm{c}^n}\;\,x is equal to
View answer and explanation
Correct option: A
2\mathrm{2}
Option C (1/2) is correct because it is derived by using the fact that A+B+C=πA + B + C = \pi and manipulating the given equation.
10 Easy
10 of 394
Understanding +4 | -1
The equation a  sin⁡  x+b  cos⁡  x=ca\,\,\sin \,\,x + b\,\,\cos \,\,x = c where ∣ c ∣  >a2+b2\left| {\,c\,} \right|\,\, > \sqrt {{a^2} + {b^2}} has
View answer and explanation
Correct option: C
No solution
Option C (No solution) is correct because the maximum value of asin⁡x+bcos⁡xa \sin x + b \cos x is a2+b2\sqrt{a^2 + b^2}, and if ∣c∣>a2+b2|c| > \sqrt{a^2 + b^2}, there is no solution.
11 Medium
11 of 394
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.
12 Easy
12 of 394
Application +4 | -1
The value of cos⁡  480∘  .  sin⁡150∘+sin⁡  600∘ . soc    390∘\cos \,\,480^\circ \,\,.\,\,\sin 150^\circ + \sin \,\,600^\circ \,.\,soc\;\,\,390^\circ is equal to
View answer and explanation
Correct option: D
−1- 1
Option D (-1) is correct because the expression simplifies to -1 after applying trigonometric identities and evaluating the values.
13 Easy
13 of 394
Analyzing +4 | -1
For all values of θ\theta, the values of 3−cos⁡  θ+cos⁡  (θ+π3)3 - \cos \,\,\theta + \cos \,\,\left( {\theta + \frac{\pi }{3}} \right) lie in the interval
View answer and explanation
Correct option: C
[2, 4]\begin{array}{llllllllllllllllllll} {\left[ {\mathrm{2, }\,\mathrm{4}} \right]} \end{array}
Option C ([2, 4]) is correct because the expression simplifies to 3 - sin(θ + π/6), and since -1 ≤ sin(θ + π/6) ≤ 1, the range is [2, 4].
14 Hard
14 of 394
Application +4 | -1
If tan⁡α=mm+1\tan \alpha=\frac{m}{m+1} and tan⁡β=12m+1\tan \beta=\frac{1}{2 m+1}, then α+β=\alpha+\beta=
View answer and explanation
Correct option: B
π4\frac{\pi}{4}
Option B (π4\frac{{\pi}}{{4}}) is correct because by using the formula for tan⁡(α+β)\tan(\alpha + \beta), we can show that tan⁡(α+β)=1\tan(\alpha + \beta) = 1, which implies α+β=π4\alpha + \beta = \frac{{\pi}}{{4}}.
15 Easy
15 of 394
Application +4 | -1
If 12  cot⁡2  θ−31    cosec    θ+32=012\,\,{\cot ^2}\,\,\theta - 31\;\,\,\mathrm{cosec}\;\,\,\theta + 32 = 0, then the value of sin⁡  θ\sin \,\,\theta is
View answer and explanation
Correct option: C
5\frac{}{5} or 34\frac{3}{4}
Option C (45\frac{4}{5} or 34\frac{3}{4}) is correct because solving the trigonometric equation 12cot⁡2θ−31cosecθ+32=012{\cot ^2}\theta - 31\mathrm{cosec}\theta + 32 = 0 leads to sin⁡θ=45\sin \theta = \frac{4}{5} or sin⁡θ=34\sin \theta = \frac{3}{4}.

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs
16 Easy
16 of 394
Understanding +4 | -1
Assertion: Area of unit circle is π\pi unit 2^2.

Reason: Radian measure of 40∘20′40^{\circ} 20^{\prime} is equal to ∣2∣π540\frac{|2| \pi}{540} radian.
View answer and explanation
Correct option: B
Assertion is correct, reason is correct; reason is not a correct explanation for assertion
Assertion is correct, reason is correct; reason is not a correct explanation for assertion because both statements are individually true, but the radian measure doesn't explain the area of a unit circle.
17 Medium
17 of 394
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.
18 Easy
18 of 394
Application +4 | -1
The value of cosec⁡(−1410∘)\operatorname{cosec}\left(-1410^{\circ}\right) is
View answer and explanation
Correct option: B
2
Option B (2) is correct because cosec⁡(−1410∘)\operatorname{cosec}(-1410^{\circ}) can be simplified using the properties of cosecant function and reference angles.
19 Easy
19 of 394
Application +4 | -1
If sin⁡x=−35\sin x=-\frac{3}{5}, where π<x<3π2\pi < x < \frac{3 \pi}{2}, then 80(tan⁡2x−cos⁡x)80\left(\tan ^2 x-\cos x\right) is equal to :
View answer and explanation
Correct option: A
109
Option A (109) is correct because given sin⁡x=−35\sin x = -\frac{3}{5} and π<x<3π2\pi < x < \frac{3\pi}{2}, we can find tan⁡x\tan x and cos⁡x\cos x, and then evaluate the expression 80(tan⁡2x−cos⁡x)80(\tan^2 x - \cos x).
20 Easy
20 of 394
Application +4 | -1
tan⁡15∘+tan⁡75∘\tan 15^{\circ}+\tan 75^{\circ} is equal to:
View answer and explanation
Correct option: D
4
Option D (4) is correct because tan⁡15∘+tan⁡75∘\tan 15^{\circ} + \tan 75^{\circ} simplifies to 4 using trigonometric identities.
21 Medium
21 of 394
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}}.
22 Medium
22 of 394
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.
23 Easy
23 of 394
Application +4 | -1
Let f(θ)=tan⁡3θ1+tan⁡2θ−cot⁡3θ1+cot⁡2θ,0<θ<π4f(\theta)=\frac{\tan ^3 \theta}{1+\tan ^2 \theta}-\frac{\cot ^3 \theta}{1+\cot ^2 \theta}, 0<\theta<\frac{\pi}{4}. Then f(θ)f(\theta) is equal to
View answer and explanation
Correct option: C
−2cot⁡(2θ)-2 \cot (2 \theta)
Option C (−2cot⁡(2θ)-2 \cot (2 \theta)) is correct because the expression simplifies to −2cot⁡(2θ)-2 \cot (2 \theta) using trigonometric identities.
24 Easy
24 of 394
Application +4 | -1
If sin⁡2x+sin⁡xcos⁡x−6cos⁡2x=0\sin ^2 x+\sin x \cos x-6 \cos ^2 x=0 and −π2<x<0\frac{-\pi}{2} < x < 0, then value of sin⁡2x\sin 2 x is
View answer and explanation
Correct option: B
−35\frac{-3}{5}
Option B (-20.5) is correct because by setting up two equations based on the sum of the first 50 terms and the sum of the next 50 terms, we can solve for the first term of the arithmetic series.
25 Medium
25 of 394
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}.
26 Hard
26 of 394
Analyzing +4 | -1
If y=sec⁡2θ  −  tan⁡θsec⁡2θ  +  tan⁡θy = \frac{{{{\sec }^2}\theta \,\, - \,\,\tan \theta }}{{{{\sec }^2}\theta \,\, + \,\,\tan \theta }}, then
View answer and explanation
Correct option: A
13<y<3\frac{1}{3} < y < 3
Option A (13<y<3\frac{1}{3} < y < 3) is correct because by substituting x=tan⁡θx = \tan \theta and analyzing the resulting quadratic equation, we find that the range of y is between 13\frac{1}{3} and 3.
27 Easy
27 of 394
Application +4 | -1
cot⁡xcot⁡2x−cot⁡2xcot⁡3x−cot⁡3xcot⁡x\cot x \cot 2 x-\cot 2 x \cot 3 x-\cot 3 x \cot x is equal to
View answer and explanation
Correct option: A
1
Option A (1) is the correct answer because the given trigonometric expression simplifies to 1 using the cotangent identity and algebraic manipulation.
28 Medium
28 of 394
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.
29 Medium
29 of 394
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.
30 Easy
30 of 394
Application +4 | -1
The value of tan⁡  1∘  tan⁡  2∘  tan⁡  3∘ …tan⁡  89∘\tan \,\,1^\circ \,\,\tan \,\,2^\circ \,\,\tan \,\,3^\circ \, \ldots \tan \,\,89^\circ is equal to
View answer and explanation
Correct option: D
1\mathrm{1}
Option D (1) is correct because the product of tan⁡θ\tan \theta and tan⁡(90°−θ)\tan (90° - \theta) is 1, and tan⁡45°=1\tan 45° = 1.
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