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JEE • MATHEMATICS • TRIGONOMETRY RATIOS

Trigonometry ratios practice · Easy

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Showing 31-60 of 218 questions
31 Easy
31 of 218
Application +4 | -1
If cos⁡248∘−sin⁡212∘=\cos ^2 48^{\circ}-\sin ^2 12^{\circ}=
View answer and explanation
Correct option: B
5+18\frac{\sqrt{5}+1}{8}
Option B (5+18\frac{{\sqrt{{5}}+1}}{{8}}) is correct because it is derived from using the identity cos⁡2A−sin⁡2B=cos⁡(A+B)cos⁡(A−B)\cos^2 A - \sin^2 B = \cos(A+B)\cos(A-B) and the value of cos⁡36∘\cos 36^{\circ}.
32 Easy
32 of 218
Application +4 | -1
If   x  \mathrm{\;}x\mathrm{\;} lies in IInd quadrant, then   1+sin⁡  x+1−sin⁡  x1+sin⁡  x−1−sin⁡  x  \;\frac{{\sqrt {1 + \sin \,\,x} + \sqrt {1 - \sin \,\,x} }}{{\sqrt {1 + \sin \,\,x} - \sqrt {1 - \sin \,\,x} }}\; is equal to
View answer and explanation
Correct option: B
tan⁡  x2\tan \,\,\frac{x}{2}
Option B is correct because by simplifying the given expression using trigonometric identities and considering the quadrant of x, the result is tan⁡ x2\tan \,\frac{x}{2}.
33 Easy
33 of 218
Application +4 | -1
A force P{P} is applied on a block placed on a rough horizontal plane. If P=20 kg{P}=20 {~kg} and weight of block is 40 kg40 {~kg}, then coefficient of friction is
View answer and explanation
Correct option: C
0.5
Option C (0.5) is correct because the coefficient of friction is calculated directly from the applied force and the weight of the block when the block is on the verge of moving.
34 Easy
34 of 218
Application +4 | -1
cot⁡215∘−1cot⁡215∘+1=\frac{\cot ^2 15^{\circ}-1}{\cot ^2 15^{\circ}+1}=
View answer and explanation
Correct option: B
32\frac{\sqrt{3}}{2}
Option B (32\frac{{\sqrt{3}}}{{2}}) is correct because after converting cotangent to cosine and sine and applying trigonometric identities, the expression simplifies to 32\frac{{\sqrt{3}}}{{2}}.
35 Easy
35 of 218
Application +4 | -1
If 2cos⁡θ+sin⁡θ=1(θ≠π2)2 \cos \theta+\sin \theta=1\left(\theta \neq \frac{\pi}{2}\right), then 7cos⁡θ+6sin⁡θ7 \cos \theta+6 \sin \theta is equal to:
View answer and explanation
Correct option: D
465\frac{46}{5}
Option D (465\frac{46}{5}) is correct because it is the result of solving the trigonometric equation and substituting the values of sinθ and cosθ.
36 Easy
36 of 218
Analyzing +4 | -1
If x=ycos⁡2π3=zcos⁡4π3x=y \cos \frac{2 \pi}{3}=z \cos \frac{4 \pi}{3}, then xy+yz+zx=x y+y z+z x= ?
View answer and explanation
Correct option: B
0
Option B (0) is correct because substituting the given relationships into the expression xy+yz+zxxy + yz + zx results in 0.
37 Easy
37 of 218
Analyzing +4 | -1
If 15sin⁡4α+10cos⁡4α=615{\sin ^4}\alpha + 10{\cos ^4}\alpha = 6, for some α∈R\alpha \in R, then the value of 27sec⁡6α+8cosec6α27{\sec ^6}\alpha + 8\mathrm{cose}{\mathrm{c}^6}\alpha is equal to
View answer and explanation
Correct option: D
250
Option D (250) is correct because by using the given trigonometric equation to find tan²α and cot²α, we can calculate the value of the expression.
38 Easy
38 of 218
Application +4 | -1
sin⁡2  17.5∘+sin⁡2  72.5∘{\sin ^2}\,\,17.5^\circ + {\sin ^2}\,\,72.5^\circ is equal to
View answer and explanation
Correct option: B
tan⁡2  45∘{\tan ^2}\,\,45^\circ
Option B (tan⁡2  45∘{\tan ^2}\,\,45^\circ) is correct because sin⁡2  17.5∘+sin⁡2  72.5∘{\sin ^2}\,\,17.5^\circ + {\sin ^2}\,\,72.5^\circ simplifies to 1, which is equal to tan⁡2  45∘{\tan ^2}\,\,45^\circ.
39 Easy
39 of 218
Application +4 | -1
If 0<x<π0 < x < \pi, and cos⁡x+sin⁡x=1/2\cos x+\sin x=1 / 2, then tan⁡x\tan x is
View answer and explanation
Correct option: D
−(4+7)/3-(4+\sqrt{7}) / 3
Option D (−(4+7)/3-(4+\sqrt{7}) / 3) is correct because it is one of the solutions to the quadratic equation in tan⁡x\tan x derived from the given equation, and it satisfies the condition 0<x<π0 < x < \pi.
40 Easy
40 of 218
Application +4 | -1
In a △ABC,∠A=π/2\triangle A B C, \angle A=\pi / 2, then cos⁡2B+cos⁡2C\cos ^2 B+\cos ^2 C equals.
View answer and explanation
Correct option: C
1
Option C (1) is correct because in a right-angled triangle, ∠A=π2\angle A = \frac{\pi}{2}, so B+C=π2B + C = \frac{\pi}{2}. Therefore, cos⁡B=sin⁡C\cos B = \sin C, and cos⁡2B+cos⁡2C=sin⁡2C+cos⁡2C=1\cos^2 B + \cos^2 C = \sin^2 C + \cos^2 C = 1.
41 Easy
41 of 218
Application +4 | -1
In a Δ   ABC,   ∠  A=π2\Delta \,\;ABC,\;\,\angle \;A = \frac{\pi }{2}, then cos⁡2  B+cos⁡2  C{\cos ^2}\,\,B + {\cos ^2}\,\,C equals

View answer and explanation
Correct option: C
1\mathrm{1}
Option C (1) is correct because in a right-angled triangle, B+C=π2B + C = \frac{\pi}{2}, and thus cos⁡2 B+cos⁡2 C=cos⁡2 B+sin⁡2 B=1{\cos ^2}\,B + {\cos ^2}\,C = {\cos ^2}\,B + {\sin ^2}\,B = 1.
42 Easy
42 of 218
Understanding +4 | -1
The equation sin⁡  x+sin⁡  y+sin⁡  z=−3   for   0≤x≤2π,   0≤y≤2π,   0≤z≤  2π\sin \,\,x + \sin \,\,y + \sin \,\,\mathrm{z} = - 3\;\,\mathrm{for}\;\,0 \le x \le 2\pi ,\;\,0 \le y \le 2\pi ,\,\;0 \le \mathrm{z} \le \,\,2\pi has

View answer and explanation
Correct option: A
One solution
Option A (One solution) is correct because the equation is only satisfied when sin(x) = sin(y) = sin(z) = -1, which occurs at x = y = z = 3π/2 within the given range.
43 Easy
43 of 218
Application +4 | -1
cot⁡2π6+cosec⁡5π6+3tan⁡2π6\cot ^2 \frac{\pi}{6}+\operatorname{cosec} \frac{5 \pi}{6}+3 \tan ^2 \frac{\pi}{6} is equal to
View answer and explanation
Correct option: A
6
Option A (6) is correct because after substituting the trigonometric values and simplifying, the expression equals 6.
44 Easy
44 of 218
Analyzing +4 | -1
The value of the expression 3(sin⁡  θ−cos⁡  θ)4+  6(sin⁡  θ+cos⁡  θ)2+  4(sin⁡6  θ+cos⁡6  θ)3{\left( {\sin \,\,\theta - \cos \,\,\theta } \right)^4} + \,\,6{\left( {\sin \,\,\theta + \cos \,\,\theta } \right)^2} + \,\,4\left( {{{\sin }^6}\,\,\theta + {{\cos }^6}\,\,\theta } \right) is
View answer and explanation
Correct option: C
13\mathrm{13}
Option C (13) is correct because expanding and simplifying the given trigonometric expression results in a constant value of 13.
45 Easy
45 of 218
Analyzing +4 | -1
If a point (x,y)=(tan⁡θ+sin⁡θ,tan⁡θ−sin⁡θ),(x,y) = (\tan \mathrm{\theta } + \sin \mathrm{\theta },\tan \mathrm{\theta } - \sin \mathrm{\theta }), then locus of (x,y)\left( {x,y} \right) is
View answer and explanation
Correct option: C
(x2−y2)2=16xy{\left( {{x^2} - {y^2}} \right)^2} = 16xy
Option C ((x2−y2)2=16xy{\left( {{x^2} - {y^2}} \right)^2} = 16xy) is correct because it is the locus of the point (x, y) when x = tan θ + sin θ and y = tan θ - sin θ.

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46 Easy
46 of 218
Analyzing +4 | -1
If π<α<3 π2\pi < \alpha < \frac{{3\,\pi }}{2}, then the expression 4sin⁡4α+sin⁡22α+4cos⁡2(π4−α2)\sqrt {4{{\sin }^4}\alpha + {{\sin }^2}2\alpha } + 4{\cos ^2}\left( {\frac{\pi }{4} - \frac{\alpha }{2}} \right) is equal to
View answer and explanation
Correct option: C
2
Option C (2) is correct because the expression simplifies to 2 using trigonometric identities and the given range of alpha.
47 Easy
47 of 218
Understanding +4 | -1
Assertion: In a unit circle, radius of circle is 1 unit.

Reason: 1 min1 \mathrm{~min} (or 1′1^{\prime}) is divided into 60 s60 \mathrm{~s}.
View answer and explanation
Correct option: B
Assertion is correct, reason is correct; reason is not a correct explanation for assertion
Option B is correct because both the assertion and the reason are true statements, but the reason does not explain why a unit circle has a radius of 1 unit.
48 Easy
48 of 218
Application +4 | -1
If   3 π4<α<π,   then   cosec2 α+2  cot⁡  α  \;\frac{{3\,\pi }}{4} < \alpha < \pi ,\,\mathrm{\;then}\;\,\sqrt {\mathrm{cose}{\mathrm{c}^2}\,\alpha + 2\,\,\cot \,\,\alpha } \mathrm{\;} is equal to

View answer and explanation
Correct option: C
−1−cot⁡  α- 1 - \cot \,\,\alpha
Option C (−1−cot⁡ α- 1 - \cot \,\alpha) is correct because it considers the range of α\alpha and simplifies the expression accordingly.
49 Easy
49 of 218
Application +4 | -1
If sin⁡  x+sin⁡2  x=1\sin \,\,x + {\sin ^2}\,\,x = 1, then the value of cos⁡12  x+3  cos⁡10  x+3  cos⁡8  x+cos⁡6  x−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: C
0\mathrm{0}
Option C (0) is correct because by substituting sin⁡x=cos⁡2x\sin x = \cos^2 x and simplifying, the expression reduces to 1 - 1 = 0.
50 Easy
50 of 218
Application +4 | -1
If cos⁡α+cos⁡β=a,sin⁡α+sin⁡β=\cos \alpha+\cos \beta=a, \sin \alpha+\sin \beta= bb and θ\theta is the arithmetic mean between α\alpha and β\beta then sin⁡2θ+cos⁡2θ\sin 2 \theta+\cos 2 \theta is equal to
View answer and explanation
Correct option: D
none of these
The correct answer is 1 because the given expression simplifies to 1 using trigonometric identities and the condition x + y = z.
51 Easy
51 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
View answer and explanation
Correct option: B
1
Option B (1) is correct because the product of tangents from 1 to 89 degrees simplifies to 1 by pairing tan⁡θ\tan \theta with tan⁡(90∘−θ)\tan (90^{\circ} - \theta).
52 Easy
52 of 218
Understanding +4 | -1
The value of cos⁡  1∘  cos⁡  2∘  cos⁡  3∘…cos⁡  179∘\cos \,\,1^\circ \,\,\cos \,\,2^\circ \,\,\cos \,\,3^\circ \ldots \cos \,\,179^\circ is
View answer and explanation
Correct option: B
0\mathrm{0}
Option B (0) is correct because the product includes cos(90°), which is 0, making the entire product zero.
53 Easy
53 of 218
Application +4 | -1
If cos⁡x+cos⁡2x=1\cos x+\cos ^2 x=1, then the value of sin⁡12x+\sin ^{12} x+ 3sin⁡10x+3sin⁡8x+sin⁡6x−13 \sin ^{10} x+3 \sin ^8 x+\sin ^6 x-1 is equal to
View answer and explanation
Correct option: D
0
Option D (0) is correct because given cos⁡x+cos⁡2x=1\cos x + \cos^2 x = 1, we can deduce cos⁡x=sin⁡2x\cos x = \sin^2 x. Substituting this into the expression and simplifying leads to 0.
54 Easy
54 of 218
Application +4 | -1
If cos⁡A=3/4\cos A=3 / 4, then the value of 16cos⁡2(A/2)−32sin⁡(A/2)sin⁡(5A/2)16 \cos ^2(A / 2)-32 \sin (A / 2) \sin (5 A / 2) is
View answer and explanation
Correct option: C
3
Option C (3) is correct because by using trigonometric identities and the given value of cos⁡A\cos A, the expression 16cos⁡2(A/2)−32sin⁡(A/2)sin⁡(5A/2)16 \cos ^2(A / 2)-32 \sin (A / 2) \sin (5 A / 2) simplifies to 3.
55 Easy
55 of 218
Application +4 | -1
sin⁡120∘cos⁡150∘−cos⁡240∘sin⁡330∘\sin 120^{\circ} \cos 150^{\circ}-\cos 240^{\circ} \sin 330^{\circ} is equal to
View answer and explanation
Correct option: B
-1
Option B (-1) is correct because substituting the values of the trigonometric functions and simplifying the expression leads to the result -1.
56 Easy
56 of 218
Understanding +4 | -1
If   x=r  sin⁡  θ  cos⁡  ϕ,   y=r  sin⁡  θ  sin⁡  ϕ  and  z=r  cos⁡  θ\;x = r\,\,\sin \,\,\mathrm{\theta }\,\,\cos \,\,\phi ,\;\,y = r\,\,\sin \,\,\mathrm{\theta }\,\,\sin \,\,\phi \mathrm{\;and\;z} = r\,\,\cos \,\,\mathrm{\theta }, then the value of x2+y2+z2{x^2} + {y^2} + {\mathrm{z}^2} is independent of
View answer and explanation
Correct option: A
θ,  ϕ\mathrm{\theta },\,\,\phi
Option A (θ,  ϕ\mathrm{\theta },\,\,\phi) is correct because x2+y2+z2{x^2} + {y^2} + {\mathrm{z}^2} simplifies to r2{r^2}, which is independent of θ\mathrm{\theta } and ϕ\phi.
57 Easy
57 of 218
Application +4 | -1
If tan⁡25∘=x\tan 25^{\circ}=x, then tan⁡155∘−tan⁡115∘1+tan⁡155∘tan⁡115∘\frac{\tan 155^{\circ}-\tan 115^{\circ}}{1+\tan 155^{\circ} \tan 115^{\circ}} is equal to
View answer and explanation
Correct option: A
1−x22x\frac{1-x^2}{2 x}
The correct answer is A (1−x22x\frac{1-x^2}{2x}) because simplifying the given expression using trigonometric identities and the given value of tan⁡25∘=x\tan 25^{\circ} = x leads to this result.
58 Easy
58 of 218
Analyzing +4 | -1
If sin⁡A+cos⁡A=m\sin A + \cos A = m and sin⁡3A+cos⁡3A=n{\sin ^3}A + {\cos ^3}A = n, then
View answer and explanation
Correct option: C
m3−3m+2n=0{m^3} - 3m + 2n = 0
Option C (m3−3m+2n=0{m^3} - 3m + 2n = 0) is correct because it represents the relationship derived by cubing the first equation and substituting the second equation.
59 Easy
59 of 218
Application +4 | -1
Let f(θ)=cot⁡θ1−tan⁡θ+tan⁡θ1−cot⁡θ,π<θ<3π4f(\theta)=\frac{\cot \theta}{1-\tan \theta}+\frac{\tan \theta}{1-\cot \theta}, \pi < \theta < \frac{3 \pi}{4}, then f(9π8)f\left(\frac{9 \pi}{8}\right) is equal to:
View answer and explanation
Correct option: B
1+221+2 \sqrt{2}
Option B (1+221+2 \sqrt{2}) is correct because after simplifying the trigonometric expression and substituting the given value, we obtain 1+221+2 \sqrt{2}.
60 Easy
60 of 218
Application +4 | -1
The maximum value of 4  sin⁡2  x  +  3  cos⁡2  x4\,\,{\sin ^2}\,\,x\,\, + \,\,3\,\,{\cos ^2}\,\,x is
View answer and explanation
Correct option: A
4\mathrm{4}
Option A (4\mathrm{4}) is correct because the maximum value of the expression is 4, which occurs when sin⁡2x=1\sin^2 x = 1.
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