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JEE • MATHEMATICS • TRIGONOMETRY RATIOS

Trigonometry ratios practice · Easy

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91 Easy
91 of 218
Application +4 | -1
If   sin⁡  Asin⁡  B=  32  and    cos⁡  Acos⁡  B=52,  0<A,  B<π2  \;\frac{{\sin \,\,A}}{{\sin \,\,B}} = \;\frac{{\sqrt 3 }}{2}\,\,\mathrm{and}\;\,\,\frac{{\cos \,\,A}}{{\cos \,\,B}} = \frac{{\sqrt 5 }}{2},\,\,0 < A,\,\,B < \frac{\pi }{2}\;, then
View answer and explanation
Correct option: A
tan⁡A=35\tan A = \frac{{\sqrt 3 }}{{\sqrt 5 }}
Option A (tan⁡A=35\tan A = \frac{{\sqrt 3 }}{{\sqrt 5 }}) is correct because, by dividing the given equations, we can directly find the value of tan⁡A\tan A.
92 Easy
92 of 218
Application +4 | -1
In a △ABC,∠A=π2\triangle A B C, \angle A=\frac{\pi}{2}, then cos⁡2B+cos⁡2C\cos ^2 B+\cos ^2 C is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because when ∠A=π2\angle A = \frac{\pi}{2}, angles B and C are complementary, so cos⁡C=sin⁡B\cos C = \sin B, and cos⁡2B+sin⁡2B=1\cos^2 B + \sin^2 B = 1.
93 Easy
93 of 218
Application +4 | -1
If 2cos⁡θ+sin⁡θ=1,(θ≠(4k+1)π/22 \cos \theta+\sin \theta=1,(\theta \neq(4 k+1) \pi / 2, k∈k \in I) then 7cos⁡θ+6sin⁡θ7 \cos \theta+6 \sin \theta is equal to:
View answer and explanation
Correct option: B
2
The correct answer is B (2) because by solving the given equation for sin θ and cos θ, we can substitute these values into the expression 7cos θ + 6sin θ to obtain the result.
94 Easy
94 of 218
Application +4 | -1
The value of sin⁡10∘+sin⁡20∘+sin⁡30∘+…+sin⁡360∘\sin 10^\circ + \sin 20^\circ + \sin 30^\circ + \ldots + \sin 360^\circ is
View answer and explanation
Correct option: B
0
Option B (0) is correct because the sines of angles equally spaced around the unit circle sum to zero due to symmetry.
95 Easy
95 of 218
Application +4 | -1
If sin⁡x+sin⁡2x=1\sin x+\sin ^2 x=1, then the value of cos⁡12x+3cos⁡10x+3cos⁡8x+cos⁡6x−1\cos ^{12} x+3 \cos ^{10} x+3 \cos ^8 x+\cos ^6 x-1 is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because by substituting sin⁡x=cos⁡2x\sin x = \cos^2 x into the given expression, it simplifies to 0.
96 Easy
96 of 218
Understanding +4 | -1
The value of   cos⁡  1∘  cos⁡  2∘  cos⁡  3∘…cos⁡  100∘  \mathrm{\;}\cos \,\,1^\circ \,\,\cos \,\,2^\circ \,\,\cos \,\,3^\circ \ldots \cos \,\,100^\circ \mathrm{\;} is equal to
View answer and explanation
Correct option: C
0\mathrm{0}
Option C (0) is correct because the product includes cos(90°), which equals 0, thus making the entire product equal to 0.
97 Easy
97 of 218
Application +4 | -1
sin⁡75∘=\sin 75^{\circ}=
View answer and explanation
Correct option: B
3+122\frac{\sqrt{3}+1}{2 \sqrt{2}}
Option B (3+122\frac{{\sqrt{{3}}+1}}{{2 \sqrt{{2}}}}) is correct because it is derived from using the sine addition formula sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A+B) = \sin A \cos B + \cos A \sin B with A = 45 and B = 30.
98 Easy
98 of 218
Application +4 | -1
If x+y=zx+y=z, then cos⁡2x+cos⁡2y+cos⁡2z−2cos⁡xcos⁡ycos⁡z\cos ^2 x+\cos ^2 y+\cos ^2 z-2 \cos x \cos y \cos z is equal to
View answer and explanation
Correct option: D
1
Option D (3+55\frac{{\sqrt{{3}} + \sqrt{{5}}}}{{\sqrt{{5}}}}) is correct because it is derived by using the given ratios to find tan⁡A\tan A and tan⁡B\tan B, then summing them.
99 Easy
99 of 218
Understanding +4 | -1
Which one of the following is possible?
View answer and explanation
Correct option: C
tan⁡  θ=45\tan \,\,\theta = 45
Option C (tan⁡ θ=45\tan \,\theta = 45) is possible because the range of the tangent function is all real numbers.
100 Easy
100 of 218
Application +4 | -1
If sin⁡θ+cosec⁡θ=2\sin \theta+\operatorname{cosec} \theta=2, the value of sin⁡10θ+cosec⁡10θ\sin ^{10} \theta+\operatorname{cosec}^{10} \theta is
View answer and explanation
Correct option: A
2
Option A (2) is correct because if sin⁡θ+cosec⁡θ=2\sin \theta + \operatorname{cosec} \theta = 2, then sin⁡θ=1\sin \theta = 1, and thus sin⁡10θ+cosec⁡10θ=110+110=2\sin^{10} \theta + \operatorname{cosec}^{10} \theta = 1^{10} + 1^{10} = 2.
101 Easy
101 of 218
Analyzing +4 | -1
All the pairs (x, y) that satisfy the inequality 2sin⁡2x−2sin⁡x+5.14sin⁡2y≤1{2^{\sqrt {{{\sin }^2}x - 2\sin x + 5} }}.\frac{1}{{{4^{{{\sin }^2}y}}}} \le 1 also satisfy the equation
View answer and explanation
Correct option: B
sin⁡x=∣sin⁡y∣\sin x = \left| {\sin y} \right|
Option B (sin⁡x=∣sin⁡y∣\sin x = \left| {\sin y} \right|) is correct because the inequality holds only when sin⁡x=1\sin x = 1 and sin⁡2y=1\sin^2 y = 1, which implies sin⁡x=∣sin⁡y∣\sin x = |\sin y|.
102 Easy
102 of 218
Application +4 | -1
The value of cosec⁡(−1410)∘\operatorname{cosec}(-1410)^{\circ} is equal to
View answer and explanation
Correct option: A
2
Option A (2) is correct because \operatorname{cosec}(-1410)° simplifies to \operatorname{cosec}(30)°, which equals 2.
103 Easy
103 of 218
Application +4 | -1
16sin⁡(20∘)sin⁡(40∘)sin⁡(80∘)16 \sin \left(20^{\circ}\right) \sin \left(40^{\circ}\right) \sin \left(80^{\circ}\right) is equal to:
View answer and explanation
Correct option: B
232 \sqrt{3}
Option B (232 \sqrt{{3}}) is correct because it uses the identity sin⁡θ⋅sin⁡(60∘−θ)⋅sin⁡(60∘+θ)=14sin⁡3θ\sin \theta \cdot \sin (60^\circ - \theta) \cdot \sin (60^\circ + \theta) = \frac{{1}}{{4}} \sin 3\theta to simplify the expression.
104 Easy
104 of 218
Analyzing +4 | -1
If x  cos⁡  α+ysin⁡  α=2a,   x  cos⁡  α+y  sin⁡  β=2ax\,\,\cos \,\,\alpha + y\sin \,\,\alpha = 2a,\;\,x\,\,\cos \,\,\alpha + y\,\,\sin \,\,\beta = 2a and 2sin⁡  α2  sin⁡  β2=12\sin \,\,\frac{\alpha }{2}\,\,\sin \,\,\frac{\beta }{2} = 1, then

View answer and explanation
Correct option: C
y2=4a  (α−x){y^2} = 4a\,\,\left( {\alpha - x} \right)
Option C (y2=4a (α−x){y^2} = 4a\,\left( {\alpha - x} \right)) is correct because it is derived from the given trigonometric equations and the condition 2sin⁡ α2 sin⁡ β2=12\sin \,\frac{\alpha }{2}\,\sin \,\frac{\beta }{2} = 1.
105 Easy
105 of 218
Application +4 | -1
If tan⁡A2=32\tan \frac{A}{2}=\frac{3}{2}, then 1+cos⁡A1−cos⁡A=\frac{1+\cos A}{1-\cos A}=
View answer and explanation
Correct option: D
4/94 / 9
Option D (49\frac{{4}}{{9}}) is correct because 1+cos⁡A1−cos⁡A\frac{{1 + \cos A}}{{1 - \cos A}} is equal to cot⁡2(A2)\cot^2(\frac{{A}}{{2}}), and since tan⁡(A2)=32\tan(\frac{{A}}{{2}}) = \frac{{3}}{{2}}, then cot⁡2(A2)=(23)2=49\cot^2(\frac{{A}}{{2}}) = (\frac{{2}}{{3}})^2 = \frac{{4}}{{9}}.

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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
106 Easy
106 of 218
Application +4 | -1
The maximum value of 1+8  sin⁡2x2  cos⁡2x21 + 8\,\,{\sin ^2}{x^2}\,\,{\cos ^2}{x^2}, is
View answer and explanation
Correct option: A
3\mathrm{3}
Option A is correct because the maximum value is found by simplifying the expression using trigonometric identities and considering the range of the cosine function.
107 Easy
107 of 218
Analyzing +4 | -1
The maximum value of cos⁡  x{cos⁡  x1−sin⁡  x  +  1−sin⁡  xcos⁡  x}\cos \,\,x\left\{ {\frac{{\cos \,\,x}}{{1 - \sin \,\,x}}\,\, + \,\,\frac{{1 - \sin \,\,x}}{{\cos \,\,x}}} \right\}, is
View answer and explanation
Correct option: C
2
Option C (2) is correct because the given expression simplifies to 2 for all valid values of x.
108 Easy
108 of 218
Application +4 | -1
cos⁡  1∘+  cos⁡  2∘+cos⁡  3∘+…+cos⁡  180∘=\cos \,\,1^\circ + \,\,\cos \,\,2^\circ + \cos \,\,3^\circ + \ldots + \cos \,\,180^\circ =
View answer and explanation
Correct option: D
−1- 1
Option D (−1-1) is correct because the sum of the cosine series from 1° to 180° simplifies to -1 due to symmetry and the value of cos(180°).
109 Easy
109 of 218
Analyzing +4 | -1
If x=∑n=0∞(−1)ntan⁡2nθx = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}{{\tan }^{2n}}\theta } and y=∑n=0∞cos⁡2nθy = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\theta } for 0<θ<π40 < \theta < \frac{\pi }{4} then
View answer and explanation
Correct option: B
y(1−x)=1y(1 - x) = 1
Option B (y(1−x)=1y(1 - x) = 1) is correct because by simplifying the given infinite sums for x and y, we can derive the relationship y(1−x)=1y(1-x) = 1.
110 Easy
110 of 218
Application +4 | -1
If sin  +cosec = 2\mathrm{sin\; + cosec}\,\mathrm{ = }\,\mathrm{2}, then sin⁡2  θ+cosec2 θ{\sin ^2}\,\,\mathrm{\theta } + \mathrm{cose}{\mathrm{c}^2}\,\mathrm{\theta } is equal to
View answer and explanation
Correct option: C
2\mathrm{2}
Option C (2) is correct because squaring the given equation and using the reciprocal identity, we find that sin⁡2 θ+cosec2 θ=2{\sin ^2}\,\theta + \mathrm{cose}{\mathrm{c}^2}\,\theta = 2.
111 Easy
111 of 218
Application +4 | -1
Let f(θ)=(1+cos⁡θ)(1+cos⁡3θ)f(\theta)=(1+\cos \theta)(1+\cos 3 \theta) (1+cos⁡5θ)(1+cos⁡7θ)(1+\cos 5 \theta)(1+\cos 7 \theta) then f(−π8)f\left(-\frac{\pi}{8}\right) is equal to:
View answer and explanation
Correct option: D
1/81 / 8
The correct answer is D (1/8) because by using the property f(-θ) = f(θ) and trigonometric identities, we can simplify the expression to obtain the result.
112 Easy
112 of 218
Application +4 | -1
If tan⁡  θ=17\tan \,\,\theta = \frac{1}{{\sqrt 7 }}, then (cosec2 θ−sec⁡2θ)(cosec2 θ+sec⁡2θ)\frac{{\left( {\mathrm{cose}{\mathrm{c}^2}\,\theta - {{\sec }^2}\theta } \right)}}{{(\mathrm{cose}{\mathrm{c}^2}\,\theta + {{\sec }^2}\theta )}} is equal to
View answer and explanation
Correct option: B
34\frac{3}{4}
Option B (34\frac{3}{4}) is correct because it is obtained by substituting the given value of tan⁡θ\tan \theta and simplifying the expression using trigonometric identities.
113 Easy
113 of 218
Analyzing +4 | -1
If L=sin⁡2(π16)−sin⁡2(π8)L = {\sin ^2}\left( {\frac{\pi }{{16}}} \right) - {\sin ^2}\left( {\frac{\pi }{8}} \right) and M=cos⁡2(π16)−sin⁡2(π8), M = {\cos ^2}\left( {\frac{\pi }{{16}}} \right) - {\sin ^2}\left( {\frac{\pi }{8}} \right),\, then
View answer and explanation
Correct option: D
M=122+12cos⁡π8M = \frac{1}{{2\sqrt 2 }} + \frac{1}{2}\cos \frac{\pi }{8}
Option D (M=122+12cos⁡π8M = \frac{1}{{2\sqrt 2 }} + \frac{1}{2}\cos \frac{\pi }{8}) is correct because it follows from applying trigonometric identities to simplify the expression for M.
114 Easy
114 of 218
Analyzing +4 | -1
For θ>π/3,\mathrm{\theta } > \mathrm{\pi }/3, the value of f(θ)=sec⁡2θ+cos⁡2θf\left( \mathrm{\theta } \right) = {\sec ^2}\mathrm{\theta } + {\cos ^2}\mathrm{\theta } always lies in the interval
View answer and explanation
Correct option: D
[2,∞)\left[ {2,\infty } \right)
Option D ([2,∞)[2,\infty)) is correct because the minimum value of the function is 2, and it can go to infinity.
115 Easy
115 of 218
Application +4 | -1
For x∈IR,  3cos⁡  (4x−5)+4x \in IR,\,\,3\cos \,\,\left( {4x - 5} \right) + 4 lies in the interval
View answer and explanation
Correct option: A
[1,7]\left[ {1,\mathrm{ }7} \right]
Option A ([1, 7]) is correct because the minimum value of 3cos⁡(4x−5)3\cos(4x-5) is -3 and the maximum value is 3, so the range of 3cos⁡(4x−5)+43\cos(4x-5) + 4 is [1, 7].
116 Easy
116 of 218
Application +4 | -1
If cos⁡  θ=817\cos \,\,\theta = \frac{8}{{17}} and θ\theta lies in the I st quadrant, then the value of cos⁡  (30∘+θ)+cos⁡  (45∘−θ)+cos⁡  (120∘−θ)\cos \,\,\left( {30^\circ + \theta } \right) + \cos \,\,\left( {45^\circ - \theta } \right) + \cos \,\,\left( {120^\circ - \theta } \right) is
View answer and explanation
Correct option: A
2317  (3−12+12)\frac{{23}}{{17}}\,\,\left( {\frac{{\sqrt 3 - 1}}{2} + \frac{1}{{\sqrt 2 }}} \right)
Option A is correct because it represents the simplified expression after applying trigonometric sum-to-product formulas and substituting the given values.
117 Easy
117 of 218
Application +4 | -1
cos⁡(540∘−θ)−sin⁡(630∘−θ)\cos \left(540^{\circ}-\theta\right)-\sin \left(630^{\circ}-\theta\right) is equal to
View answer and explanation
Correct option: A
0
The correct answer is A (0) because cos⁡(540∘−θ)−sin⁡(630∘−θ)\cos(540^{\circ} - \theta) - \sin(630^{\circ} - \theta) simplifies to −cos⁡θ−(−cos⁡θ)=0-\cos \theta - (-\cos \theta) = 0.
118 Easy
118 of 218
Application +4 | -1
The value of 6(sin⁡6θ+cos⁡6θ)−9  (sin⁡4θ+cos⁡4θ)+4  6({\sin ^6}\mathrm{\theta } + {\cos ^6}\mathrm{\theta }) - 9\mathrm{\;}({\sin ^4}\mathrm{\theta } + {\cos ^4}\mathrm{\theta }) + 4\mathrm{\;} is
View answer and explanation
Correct option: C
1
Option C (1) is correct because the expression simplifies to 1 using trigonometric identities.
119 Easy
119 of 218
Analyzing +4 | -1
If (sec⁡A−tan⁡A)  (sec⁡B−tan⁡   B)  (sec⁡C−tan⁡C)=(sec⁡A+tan⁡A)  (sec⁡B+tan⁡B)  (sec⁡C+tan⁡C)\left( {\sec A - \tan A} \right)\,\,\left( {\sec B - \tan \;\,B} \right)\,\,\left( {\sec C - \tan C} \right) = \left( {\sec A + \tan A} \right)\,\,\left( {\sec B + \tan B} \right)\,\,\left( {\sec C + \tan C} \right) then each side is equal to
View answer and explanation
Correct option: D
±1\pm 1
Option D (±1\pm 1) is correct because the given equation implies that each side is equal to either 1 or -1.
120 Easy
120 of 218
Application +4 | -1
Let fk(x)=1k(sin⁡kx+cos⁡kx){f_k}\left( x \right) = \frac{1}{k}\left( {{{\sin }^k}x + {{\cos }^k}x} \right) for k=1,2,3...k = 1,2,3... Then, for all x∈Rx \in R , the value of f4(x)−f6(x){f_4}\left( x \right) - {f_6}\left( x \right) is equal to
View answer and explanation
Correct option: A
112\frac{1}{{12}}
Option A (112\frac{1}{{12}}) is correct because it's the result of simplifying the expression f4(x)−f6(x){f_4}(x) - {f_6}(x) using trigonometric identities.
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