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JEE • MATHEMATICS • TRIGONOMETRY RATIOS

Trigonometry ratios practice · Easy

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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 61-90 of 218 questions
61 Easy
61 of 218
Application +4 | -1
If α,β,γ\alpha, \beta, \gamma are three distinct real numbers such that 0<α,β,γ<π/20<\alpha, \beta, \gamma<\pi / 2, then tan⁡(α−β)+tan⁡(β−γ)+\tan (\alpha-\beta)+\tan (\beta-\gamma)+ tan⁡(γ−α)\tan (\gamma-\alpha) is equal to:
View answer and explanation
Correct option: D
none of these
The correct answer is D (none of these) because the given expression simplifies to the product of the tangents, not a constant value.
62 Easy
62 of 218
Application +4 | -1
If sec⁡  θ=m\sec \,\,\theta = m and tan⁡  θ=n\tan \,\,\theta = n, then 1m  [(m+n)+1(m+n)]\frac{1}{m}\,\,\left[ {\left( {m + n} \right) + \frac{1}{{\left( {m + n} \right)}}} \right] is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the expression simplifies to 2 using trigonometric identities.
63 Easy
63 of 218
Application +4 | -1
If tan⁡θ+sec⁡θ=ex\tan \theta+\sec \theta=e^x, then cos⁡θ\cos \theta equals
View answer and explanation
Correct option: B
2(ex+e−x)\frac{2}{\left(e^x+e^{-x}\right)}
Option B (2(ex+e−x)\frac{{2}}{{\left(e^x+e^{-x}\right)}}) is correct because by using the identity sec⁡2θ−tan⁡2θ=1\sec^2 \theta - \tan^2 \theta = 1, we can find cos⁡θ\cos \theta.
64 Easy
64 of 218
Application +4 | -1
If cos⁡(θ−α)=a,sin⁡(θ−β)=b\cos (\theta-\alpha)=a, \sin (\theta-\beta)=b, then cos⁡2(α−β)+2absin⁡(α−β)\cos ^2(\alpha-\beta)+2 a b \sin (\alpha-\beta) is equal to
View answer and explanation
Correct option: B
a2+b2a^2+b^2
Option B (a2+b2a^2+b^2) is correct because using trigonometric identities, the given expression simplifies to a2+b2a^2 + b^2.
65 Easy
65 of 218
Application +4 | -1
θ\theta eliminant between the equations x=asec⁡3θ,y=btan⁡3θx=a \sec ^3 \theta, y=b \tan ^3 \theta is
View answer and explanation
Correct option: B
(xa)2/3−(yb)2/3=1\left(\frac{x}{a}\right)^{2 / 3}-\left(\frac{y}{b}\right)^{2 / 3}=1
Option B ((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 sec⁡θ\sec \theta and tan⁡θ\tan \theta in terms of x, y, a, and b, and then using the identity sec⁡2θ−tan⁡2θ=1\sec^2 \theta - \tan^2 \theta = 1.
66 Easy
66 of 218
Application +4 | -1
If A+B+C=(2n+1)πA+B+C=(2 n+1) \pi, then tan⁡A+tan⁡B+tan⁡C\tan A+\tan B+\tan C is equal to
View answer and explanation
Correct option: A
tan⁡Atan⁡Btan⁡C\tan A \tan B \tan C
Option A (tan⁡Atan⁡Btan⁡C\tan A \tan B \tan C) is correct because when A+B+C=(2n+1)πA+B+C=(2n+1)\pi, the identity tan⁡A+tan⁡B+tan⁡C=tan⁡Atan⁡Btan⁡C\tan A + \tan B + \tan C = \tan A \tan B \tan C holds true.
67 Easy
67 of 218
Application +4 | -1
In a △ABC\triangle A B C, if a,ba, b and AA are given, then there are two triangles with third sides c1c_1 and c2c_2, then (c1−c2)\left(c_1-c_2\right) is
View answer and explanation
Correct option: C
2a2−b2sin⁡2A2 \sqrt{a^2-b^2 \sin ^2 A}
Option C (2a2−b2sin⁡2A2 \sqrt{a^2-b^2 \sin ^2 A}) is correct because it is derived from solving the quadratic equation for the third side cc using the Law of Cosines and then finding the difference between the two possible solutions.
68 Easy
68 of 218
Application +4 | -1
In any △ABC,4△(cot⁡A+cot⁡B+cot⁡C)\triangle A B C, 4 \triangle(\cot A+\cot B+\cot C) is equal to
View answer and explanation
Correct option: C
a2+b2+c2a^2+b^2+c^2
Option C (a2+b2+c2a^2+b^2+c^2) is correct because the expression 4△(cot⁡A+cot⁡B+cot⁡C)4 \triangle(\cot A+\cot B+\cot C) simplifies to a2+b2+c2a^2+b^2+c^2 using trigonometric identities and area formulas.
69 Easy
69 of 218
Application +4 | -1
The value of

cos⁡  (270∘+θ)  cos⁡  (90∘−θ)−sin⁡  (270∘−θ)  cos⁡  θ\cos \,\,\left( {270^\circ + \theta } \right)\,\,\cos \,\,\left( {90^\circ - \theta } \right) - \sin \,\,\left( {270^\circ - \theta } \right)\,\,\cos \,\,\theta is
View answer and explanation
Correct option: D
1\mathrm{1}
Option D is correct because simplifying the trigonometric expression using trigonometric identities results in 1.
70 Easy
70 of 218
Application +4 | -1
cos⁡11∘−sin⁡11∘cos⁡11∘+sin⁡11∘\frac{\cos 11^{\circ}-\sin 11^{\circ}}{\cos 11^{\circ}+\sin 11^{\circ}} is equal to:
View answer and explanation
Correct option: A
cot⁡56∘\cot 56^{\circ}
Option A (cot⁡56∘\cot 56^{\circ}) is correct because the expression simplifies to cot⁡56∘\cot 56^{\circ} by dividing by cos⁡11∘\cos 11^{\circ} and using the tangent subtraction formula.
71 Easy
71 of 218
Application +4 | -1
If cos⁡A3=cos⁡B4=15,−π/2<A,B<0\frac{\cos A}{3}=\frac{\cos B}{4}=\frac{1}{5},-\pi / 2< A, B < 0, then value of 2sin⁡A+4sin⁡B2 \sin A+4 \sin B is equal to
View answer and explanation
Correct option: C
-4
Option C (-4) is correct because by finding sinA and sinB using the given information and substituting them into the expression, the result is -4.
72 Easy
72 of 218
Application +4 | -1
If tan⁡θ=1/3\tan \theta=1 / \sqrt{3}, then the value of 3cos⁡2θ+sin⁡2θ\sqrt{3} \cos 2 \theta+\sin 2 \theta is
View answer and explanation
Correct option: A
3\sqrt{3}
Option A (3\sqrt{3}) is correct because substituting tan⁡θ=1/3\tan \theta = 1/\sqrt{3} into the expression 3cos⁡2θ+sin⁡2θ\sqrt{3} \cos 2 \theta+\sin 2 \theta yields 3\sqrt{3}.
73 Easy
73 of 218
Application +4 | -1
In tan⁡  θ+sec⁡  θ=3,   0<θ<π\tan \,\,\theta + \sec \,\,\theta = \sqrt 3 ,\;\,0 < \theta < \pi, then θ\theta is equal to

View answer and explanation
Correct option: C
π/6\pi /6
Option C (π/6\pi /6) is correct because solving the given equation tan⁡θ+sec⁡θ=3\tan \theta + \sec \theta = \sqrt 3 leads to this value within the specified range.
74 Easy
74 of 218
Application +4 | -1
If A=130∘A=130^{\circ} and x=sin⁡A+cos⁡Ax=\sin A+\cos A, then
View answer and explanation
Correct option: A
x>0x > 0
The correct answer is A (x>0x > 0) because when A=130∘A = 130^{\circ}, x=sin⁡A+cos⁡A=sin⁡130∘+cos⁡130∘=sin⁡50∘−cos⁡50∘x = \sin A + \cos A = \sin 130^{\circ} + \cos 130^{\circ} = \sin 50^{\circ} - \cos 50^{\circ}, and since sin⁡50∘>∣cos⁡50∘∣\sin 50^{\circ} > |\cos 50^{\circ}|, xx is positive.
75 Easy
75 of 218
Application +4 | -1
If acos⁡A−bsin⁡A=ca \cos A-b \sin A=c, then asin⁡A+bcos⁡Aa \sin A+b \cos A is equal to
View answer and explanation
Correct option: A
±a2+b2−c2\pm \sqrt{a^2+b^2-c^2}
Option A (±a2+b2−c2\pm \sqrt{{a^2+b^2-c^2}}) is correct because it is derived by squaring both the given equation and the expression to be found, then adding them.

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76 Easy
76 of 218
Application +4 | -1
The equation 8sec⁡2θ−6sec⁡θ+1=08 \sec ^2 \theta-6 \sec \theta+1=0 has -
View answer and explanation
Correct option: D
No roots
Option D (No roots) is correct because the equation leads to sec⁡θ\sec \theta values that are not possible, as the absolute value of sec⁡θ\sec \theta must be greater than or equal to 1.
77 Easy
77 of 218
Application +4 | -1
If f(x)=cos⁡2  x+sec⁡2  xf\left( x \right) = {\cos ^2}\,\,x + {\sec ^2}\,\,x, its value always is
View answer and explanation
Correct option: B
f(x)=1f\left( x \right) = 1
Option D (f(x)≥2f(x) \ge 2) is correct because the minimum value of cos⁡2x+sec⁡2x{\cos ^2}x + {\sec ^2}x occurs when cos⁡2x=1\cos^2 x = 1 and sec⁡2x=1\sec^2 x = 1, so the minimum value is 2.
78 Easy
78 of 218
Analyzing +4 | -1
Let nn be a positive integer such that sin⁡  π2n+cos⁡  π2n=n2\sin \,\,\frac{\pi }{{2n}} + \cos \,\,\frac{\pi }{{2n}} = \frac{{\sqrt n }}{2}. Then
View answer and explanation
Correct option: A
n=6n = 6
Option A (n=6n = 6) is correct because it is the only integer solution that satisfies the given trigonometric equation after squaring and simplifying.
79 Easy
79 of 218
Application +4 | -1
The maximum value of cos⁡2(π3−x)  −cos⁡2(π3+x){\cos ^2}\left( {\frac{\pi }{3} - x} \right)\,\, - {\cos ^2}\left( {\frac{\pi }{3} + x} \right) is
View answer and explanation
Correct option: C
32\frac{{\sqrt 3 }}{2}
Option C is correct because the maximum value is found by simplifying the expression and using the maximum value of the sine function.
80 Easy
80 of 218
Application +4 | -1
log⁡tan⁡1∘+log⁡tan⁡2∘+…..+log⁡tan⁡89∘\log \tan 1^{\circ}+\log \tan 2^{\circ}+\ldots . .+\log \tan 89^{\circ} is
View answer and explanation
Correct option: B
0
Option B (0) is correct because log⁡tan⁡1∘+log⁡tan⁡2∘+⋯+log⁡tan⁡89∘=log⁡(tan⁡1∘⋅tan⁡2∘⋅⋯⋅tan⁡89∘)\log \tan 1^{\circ} + \log \tan 2^{\circ} + \dots + \log \tan 89^{\circ} = \log (\tan 1^{\circ} \cdot \tan 2^{\circ} \cdot \dots \cdot \tan 89^{\circ}). Since tan⁡θ⋅tan⁡(90∘−θ)=1\tan \theta \cdot \tan (90^{\circ} - \theta) = 1, the product inside the logarithm is 1, and log⁡1=0\log 1 = 0.
81 Easy
81 of 218
Application +4 | -1
If tan⁡A=12\tan A=\frac{1}{2}, then tan⁡3A=\tan 3 A=
View answer and explanation
Correct option: B
112\frac{11}{2}
Option B (112\frac{{11}}{{2}}) is correct because using the triple angle formula for tangent, tan⁡3A\tan 3A is calculated as 112\frac{{11}}{{2}}.
82 Easy
82 of 218
Application +4 | -1
If sin⁡x  cos⁡x  cos⁡  2x=λ\sin x\,\,\cos x\,\,\cos \,\,2x = \lambda has a solution, then λ\lambda lies in the interval
View answer and explanation
Correct option: A
[−1/4,   1/4]\left[ { - 1/4,\;\,1/4} \right]
Option A is correct because the range of λ\lambda is found by considering the range of the sine function.
83 Easy
83 of 218
Application +4 | -1
If sin⁡  x+sin⁡2  x=1\sin \,\,x + {\sin ^2}\,\,x = 1, then cos⁡12  x+3  cos⁡10  x+3  cos⁡8  x+cos⁡6  x{\cos ^{12}}\,\,x + 3\,\,{\cos ^{10}}\,\,x + 3\,\,{\cos ^8}\,\,x + {\cos ^6}\,\,x is equal to
View answer and explanation
Correct option: A
1\mathrm{1}
Option B (tan⁡ B−tan⁡ C−tan⁡ A\tan \,B - \tan \,C - \tan \,A) is correct because it is derived by applying the tangent addition formula and rearranging the terms.
84 Easy
84 of 218
Application +4 | -1
If tan⁡x=ba\tan x=\frac{b}{a} then a+ba−b+a−ba+b\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}} is equal to
View answer and explanation
Correct option: B
2cos⁡xcos⁡2x\frac{2 \cos x}{\sqrt{\cos 2 x}}
Option B (2cos⁡xcos⁡2x\frac{{2 \cos x}}{{\sqrt{{\cos 2 x}}}}) is correct because the expression simplifies to this form after algebraic manipulation and using trigonometric identities.
85 Easy
85 of 218
Understanding +4 | -1
Consider the statements given below:

I. 2cos⁡x⋅cos⁡y=cos⁡(x+y)−cos⁡(x−y)2 \cos x \cdot \cos y=\cos (x+y)-\cos (x-y).

II. −2sin⁡x⋅sin⁡y=cos⁡(x+y)−cos⁡(x−y)-2 \sin x \cdot \sin y=\cos (x+y)-\cos (x-y).

III. 2sin⁡x⋅cos⁡y=sin⁡(x+y)−sin⁡(x−y)2 \sin x \cdot \cos y=\sin (x+y)-\sin (x-y).

IV. 2cos⁡x⋅sin⁡y=sin⁡(x+y)+sin⁡(x−y)2 \cos x \cdot \sin y=\sin (x+y)+\sin (x-y).

Choose the correct statements.
View answer and explanation
Correct option: B
II is correct
Option B is correct because the identity −2sin⁡x⋅sin⁡y=cos⁡(x+y)−cos⁡(x−y)-2 \sin x \cdot \sin y = \cos(x+y) - \cos(x-y) is a standard trigonometric identity.
86 Easy
86 of 218
Application +4 | -1
The least value of cosec⁡2x+25sec⁡2x\operatorname{cosec}^2 x+25 \sec ^2 x is
View answer and explanation
Correct option: D
36
Option D (36) is correct because it is the minimum value obtained by completing the square and using the AM-GM inequality.
87 Easy
87 of 218
Application +4 | -1
The greatest and least value of sin⁡  x  cos⁡  x\sin \,\,x\,\,\cos \,\,x are respectively
View answer and explanation
Correct option: B
12,  −12\frac{1}{2},\,\, - \frac{1}{2}
Option B (12,  −12\frac{1}{2},\,\, - \frac{1}{2}) is correct because it represents the maximum and minimum values of the function 12sin⁡2x\frac{1}{2}\sin 2x.
88 Easy
88 of 218
Application +4 | -1
If A=130∘A = 130^\circ and x=sin⁡A+cos⁡Ax = \sin A + \cos A, then
View answer and explanation
Correct option: A
x>0x > 0
Option A (x>0x > 0) is correct because sin⁡130∘\sin 130^\circ is positive and greater than the absolute value of cos⁡130∘\cos 130^\circ, which is negative.
89 Easy
89 of 218
Application +4 | -1
If the value of 3cos⁡36∘+5sin⁡18∘5cos⁡36∘−3sin⁡18∘\frac{3 \cos 36^{\circ}+5 \sin 18^{\circ}}{5 \cos 36^{\circ}-3 \sin 18^{\circ}} is a5−bc,\frac{\mathrm{a} \sqrt{5}-\mathrm{b}}{\mathrm{c}}, where a,b,c\mathrm{a}, \mathrm{b}, \mathrm{c} are natural numbers and gcd⁡(a,c)=1,\operatorname{gcd}(\mathrm{a}, \mathrm{c})=1, then a+b+c\mathrm{a}+\mathrm{b}+\mathrm{c} is equal to:
View answer and explanation
Correct option: C
52
Option C (52) is correct because after simplifying the given expression using trigonometric values and rationalizing the denominator, the values of a, b, and c can be determined, and their sum is 52.
90 Easy
90 of 218
Analyzing +4 | -1
If 0<x<π2,  0 < x < \frac{\pi }{2},\; then minimum value of cos3xsin⁡x+sin⁡3xcos⁡x  \frac{{\mathrm{co}{\mathrm{s}^3}x}}{{\sin x}} + \frac{{{{\sin }^3}x}}{{\cos x}}\; is
View answer and explanation
Correct option: D
1
The correct answer is D. Using AM-GM inequality, we can find that the minimum value of the expression is 1.
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