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JEE • MATHEMATICS • TRIGONOMETRY RATIOS

Trigonometry ratios practice · Easy

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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 218 questions
1 Easy
1 of 218
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.
2 Easy
2 of 218
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.
3 Easy
3 of 218
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.
4 Easy
4 of 218
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.
5 Easy
5 of 218
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.
6 Easy
6 of 218
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.
7 Easy
7 of 218
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].
8 Easy
8 of 218
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}.
9 Easy
9 of 218
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.
10 Easy
10 of 218
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.
11 Easy
11 of 218
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).
12 Easy
12 of 218
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.
13 Easy
13 of 218
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.
14 Easy
14 of 218
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.
15 Easy
15 of 218
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.

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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 218
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.
17 Easy
17 of 218
Application +4 | -1
If sin⁡θ+cos⁡θ=1\sin \theta+\cos \theta=1, then sin⁡θcos⁡θ=\sin \theta \cos \theta=
View answer and explanation
Correct option: A
0
Option A (0) is correct because squaring the given equation and using the identity \sin2 θ + \cos2 θ = 1 leads to \sin θ \cos θ = 0.
18 Easy
18 of 218
Application +4 | -1
Let f(θ)=tan⁡(π3+θ)tan⁡(π3−θ)f(\theta)=\tan \left(\frac{\pi}{3}+\theta\right) \tan \left(\frac{\pi}{3}-\theta\right), −π3<θ<π3-\frac{\pi}{3}<\theta<\frac{\pi}{3}, then f(π12)f\left(\frac{\pi}{12}\right) is equal to:
View answer and explanation
Correct option: A
2+32+\sqrt{3}
The correct answer is A (2+32+\sqrt{3}) because by using trigonometric identities and substituting the value of θ\theta, we can simplify the expression to obtain the result.
19 Easy
19 of 218
Application +4 | -1
The value of cos⁡15∘−sin⁡15∘\cos 15^{\circ}-\sin 15^{\circ} is equal to
View answer and explanation
Correct option: A
12\frac{1}{\sqrt{2}}
Option A (12\frac{{1}}{{\sqrt{{2}}}}) is correct because cos⁡15∘−sin⁡15∘\cos 15^{\circ} - \sin 15^{\circ} can be simplified to sin⁡75∘−sin⁡15∘\sin 75^{\circ} - \sin 15^{\circ}, which then simplifies to 12\frac{{1}}{{\sqrt{{2}}}}.
20 Easy
20 of 218
Application +4 | -1
If sin⁡  x+sin⁡2  x=1\sin \,\,x + {\sin ^2}\,\,x = 1, then cos⁡8  x+2cos⁡6  x+cos⁡4  x={\cos ^8}\,\,x + 2{\cos ^6}\,\,x + {\cos ^4}\,\,x =
View answer and explanation
Correct option: D
1\mathrm{1}
Option D (1) is correct because by substituting sin⁡x=cos⁡2x\sin x = \cos^2 x into the expression cos⁡8x+2cos⁡6x+cos⁡4x{\cos ^8}\\ x + 2{\cos ^6}\\ x + {\cos ^4}\\ x, we can simplify it to (sin⁡x+sin⁡2x)2(\sin x + \sin^2 x)^2, which equals 1.
21 Easy
21 of 218
Application +4 | -1
If 0<θ<π,0 < \mathrm{\theta } < \mathrm{\pi }, then the minimum value of sin⁡5θ+cosec5θ{\sin ^5}\theta + \mathrm{cose}{\mathrm{c}^5}\mathrm{\theta } is
View answer and explanation
Correct option: C
2
Option C (2) is correct because the minimum value is found using the fact that x+1x≥2x + \frac{1}{x} \ge 2 for all x>0x > 0, applied to the given expression.
22 Easy
22 of 218
Application +4 | -1
The value of (1+cos⁡π6)(1+cos⁡π3)(1+cos⁡2π3)(1+cos⁡7π6)\left(1+\cos \frac{\pi}{6}\right)\left(1+\cos \frac{\pi}{3}\right)\left(1+\cos \frac{2 \pi}{3}\right)\left(1+\cos \frac{7 \pi}{6}\right) is m16\frac{m}{16}. Value of mm is
View answer and explanation
Correct option: A
3
Option A (6) is correct because given \cos θ = −35\frac{{-3}}{{5}} and θ in the third quadrant, calculating the expression results in 16\frac{{1}}{{6}}.
23 Easy
23 of 218
Application +4 | -1
The value of sin⁡  (π+θ)  sin⁡  (π−θ)  cosec2 θ\sin \,\,\left( {\pi + \theta } \right)\,\,\sin \,\,\left( {\pi - \theta } \right)\;\mathrm{cose}{\mathrm{c}^2}\,\theta is equal to
View answer and explanation
Correct option: A
&nbsp;- 1
Option A (- 1) is correct because the expression simplifies to -1 using trigonometric identities.
24 Easy
24 of 218
Application +4 | -1
If   sin⁡x+cosec  x=2    \mathrm{\;}\sin x + \mathrm{cosec\;}x = 2\mathrm{\;\;} then sin⁡nx+cosecnx  {\sin ^n}x + \mathrm{cose}{\mathrm{c}^n}x\; is equal to
View answer and explanation
Correct option: A
2
Option A (2) is correct because if sin x + cosec x = 2, then sin x = 1, and therefore sin^n x + cosec^n x = 1 + 1 = 2.
25 Easy
25 of 218
Application +4 | -1
If x=acos⁡3θ,y=bsin⁡3θx=a \cos ^3 \theta, y=b \sin ^3 \theta, then
View answer and explanation
Correct option: C
(xa)2/3+(yb)2/3=1\left(\frac{x}{a}\right)^{2 / 3}+\left(\frac{y}{b}\right)^{2 / 3}=1
Option C ((xa)2/3+(yb)2/3=1\left(\frac{{x}}{{a}}\right)^{2 / 3}+\left(\frac{{y}}{{b}}\right)^{2 / 3}=1) is correct because it is derived by expressing cos⁡θ\cos \theta and sin⁡θ\sin \theta in terms of x, y, a, and b, and then using the identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1.
26 Easy
26 of 218
Application +4 | -1
If in a triangle ABC,2cot⁡Bcot⁡C=1A B C, 2 \cot B \cot C=1, then cos⁡Bcos⁡C\cos B \cos C will be equal to
View answer and explanation
Correct option: C
cos⁡A\cos A
Option C (cos⁡A\cos A) is correct because the given condition 2cot⁡Bcot⁡C=12 \cot B \cot C = 1 can be manipulated using trigonometric identities to show that cos⁡Bcos⁡C=cos⁡A\cos B \cos C = \cos A.
27 Easy
27 of 218
Application +4 | -1
If A, B, C, D\mathrm{A,}\,\mathrm{B,}\,\mathrm{C,}\,\mathrm{D} are the angles of a cyclic quadrilateral, then cos⁡  A+cos⁡  B+cos⁡  D  \cos \,\,A + \cos \,\,B + \cos \,\,D\; is equal to
View answer and explanation
Correct option: D
0\mathrm{0}
Option D (0) is correct because in a cyclic quadrilateral, opposite angles are supplementary, leading to cos⁡A+cos⁡C=0\cos A + \cos C = 0 and cos⁡B+cos⁡D=0\cos B + \cos D = 0, thus the sum is 0.
28 Easy
28 of 218
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 value of 2sin⁡A+4sin⁡B2 \sin A+4 \sin B is −a-a. The value of &#39;aa&#39; is
View answer and explanation
Correct option: A
4
Option A (4) is correct because calculating 2\sin A + 4\sin B with the given conditions results in -4, and since it's equal to -a, a = 4.
29 Easy
29 of 218
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\mathrm{1}
Option D (1) is correct because by expanding and simplifying 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 }, we can use the identity sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 to arrive at 1.
30 Easy
30 of 218
Application +4 | -1
If α+β=π2\alpha+\beta=\frac{\pi}{2} and β+γ=α\beta+\gamma=\alpha, then tan⁡α=\tan \alpha=
View answer and explanation
Correct option: B
tan⁡β+2tan⁡γ\tan \beta+2 \tan \gamma
Option B (tan⁡β+2tan⁡γ\tan \beta+2 \tan \gamma) is correct because it follows from trigonometric identities and algebraic manipulation of the given equations.
Revision notes and key takeaways

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Key Takeaways

  • ✓ This set contains 218 single-correct-answer MCQs on TRIGONOMETRY RATIOS.
  • ✓ Try each question before opening its answer. Read the explanation where one is available.
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