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JEE • MATHEMATICS • TRIGONOMETRY RATIOS

Trigonometry ratios practice · Mixed difficulty

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Showing 61-90 of 394 questions

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61 Medium
61 of 394
Analyzing +4 | -1
Let f(x)=1+2cos⁡x+3sin⁡xf(x)=1+2 \cos x+3 \sin x. If αf(x)+βf(x−γ)=1\alpha f(x)+\beta f(x-\gamma)=1 where α,β,γ∈R\alpha, \beta, \gamma \in R holds for all x∈Rx \in R, then βcos⁡γα\frac{\beta \cos \gamma}{\alpha} is equal to
View answer and explanation
Correct option: A
-1
Option A (-1) is correct because by finding f(x-π) and comparing αf(x) + βf(x-γ) = 1, we get α = β = 1/2 and γ = π, thus βcosγ/α = -1.
62 Easy
62 of 394
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.
63 Medium
63 of 394
Application +4 | -1
If x=acos⁡2θsin⁡θx=a \cos ^2 \theta \sin \theta and y=y= a sin⁡2θcos⁡θ\sin ^2 \theta \cos \theta, then (x2+y2)3\left(x^2+y^2\right)^3 is
View answer and explanation
Correct option: B
a2x2y2a^2 x^2 y^2
Option B (a2x2y2a^2 x^2 y^2) is correct because by substituting the given expressions for x and y, and simplifying, we find that (x2+y2)3=a2x2y2(x^2+y^2)^3 = a^2 x^2 y^2.
64 Medium
64 of 394
Application +4 | -1
If θ\theta and ϕ\phi are acute angles such that sin⁡θ=1/2\sin \theta=1 / 2 and cos⁡ϕ=1/3\cos \phi=1 / 3, then θ+ϕ\theta+\phi lies is-
View answer and explanation
Correct option: B
π/2,2π/3\pi / 2,2 \pi / 3
Option B (π/2,2π/3\pi / 2,2 \pi / 3) is correct because given the acute angles and their respective sine and cosine values, the sum of the angles lies within this interval.
65 Easy
65 of 394
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}}.
66 Easy
66 of 394
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.
67 Medium
67 of 394
Application +4 | -1
sin⁡36∘sin⁡72∘sin⁡108∘sin⁡144∘=\sin 36^{\circ} \sin 72^{\circ} \sin 108^{\circ} \sin 144^{\circ}=
View answer and explanation
Correct option: D
5/165 / 16
Option D (5/165 / 16) is correct because after applying trigonometric identities and using known values, the expression simplifies to 5/165/16.
68 Easy
68 of 394
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.
69 Medium
69 of 394
Application +4 | -1
In △ABC\triangle A B C, if sin⁡2A+sin⁡2B=sin⁡2C\sin ^{2} A+\sin ^{2} B=\sin ^{2} C, then ∠C=\angle C=
View answer and explanation
Correct option: D
90∘90^{\circ}
Option D (90°) is correct because the given equation sin2(A) + sin2(B) = sin2(C) implies a2 + b2 = c2, which is the Pythagorean theorem, indicating that angle C is 90 degrees.
70 Medium
70 of 394
Application +4 | -1
If Tn=cos⁡n  θ+sin⁡n  θ{T_n} = {\cos ^n}\,\,\theta + {\sin ^n}\,\,\theta, then 2 T6−3 T4+1=2\,{T_6} - 3\,{T_4} + 1 =
View answer and explanation
Correct option: C
0\mathrm{0}
Option C (0) is correct because the expression 2T_6 - 3T_4 + 1 simplifies to 0 using trigonometric identities and the given definition of T_n.
71 Easy
71 of 394
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.
72 Easy
72 of 394
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.
73 Hard
73 of 394
Application +4 | -1
If the roots of x3−7x2+5=0x^3-7 x^2+5=0 are tan⁡α,tan⁡β\tan \alpha, \tan \beta and tan⁡γ\tan \gamma then find the value of sec⁡2αsec⁡2βsec⁡2γ\sec ^2 \alpha \sec ^2 \beta \sec ^2 \gamma.
View answer and explanation
Correct option: A
145
Option A (145) is correct because by transforming the given cubic equation into one with roots sec⁡2α\sec^2 \alpha, sec⁡2β\sec^2 \beta, and sec⁡2γ\sec^2 \gamma, the product of the roots of the new equation is 145.
74 Easy
74 of 394
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.
75 Medium
75 of 394
Application +4 | -1
If   sin  θ+cos⁡  θ=m    and  sec⁡  θ+cosec   θ=n,  then    n(m+1)  (m−1)\mathrm{\;sin}\,\,\mathrm{\theta } + \cos \,\,\mathrm{\theta } = m\mathrm{\;}\,\,\mathrm{and}\,\,\sec \,\,\mathrm{\theta } + \mathrm{cosec}\,\mathrm{\;\theta } = n,\mathrm{\;then\;}\,\,n\left( {m + 1} \right)\,\,\left( {m - 1} \right) is equal to

View answer and explanation
Correct option: C
2m2m
Option C (2m2m) is correct because simplifying the given expression using trigonometric identities leads to the result 2m2m.

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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
76 Easy
76 of 394
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.
77 Medium
77 of 394
Application +4 | -1
One root of the equation cos⁡  θ−θ+12=0\cos \,\,\theta - \theta + \frac{1}{2} = 0 lies in the interval
View answer and explanation
Correct option: A
(0,   π/2)\left( {0,\,\;\pi /2} \right)
Option A ((0,  π/2)\left( {0,\;\pi /2} \right)) is correct because the function changes sign within this interval, indicating the presence of a root.
78 Medium
78 of 394
Analyzing +4 | -1
If A=sin⁡2θ+cos⁡4θA=\sin ^2 \theta+\cos ^4 \theta, then for all values of θ\theta
View answer and explanation
Correct option: B
34≤A≤1\frac{3}{4} \leq A \leq 1
Option B (34≤A≤1\frac{{3}}{{4}} \leq A \leq 1) is correct because the expression can be rewritten in terms of sin⁡22θ\sin^2 2\theta and its range can be determined.
79 Medium
79 of 394
Application +4 | -1
If A=130∘A=130^{\circ} and x=sin⁡A+cos⁡Ax=\sin A+\cos \mathrm{A}, then
View answer and explanation
Correct option: A
x>0x > 0
Option A (x>0x > 0) is correct because sin⁡A\sin A is positive and cos⁡A\cos A is negative, but sin⁡A\sin A has a larger magnitude, resulting in a positive sum.
80 Medium
80 of 394
Analyzing +4 | -1
If xacos⁡  α+yb  sin⁡α=1,xacos⁡β+yb  sin⁡β=1\frac{x}{a}\cos \,\,\alpha + \frac{y}{b}\,\,\sin \alpha = 1,\frac{x}{a}\cos \beta + \frac{y}{b}\,\,\sin \beta = 1 and cos⁡α  cos⁡βa2+sin⁡α  sin⁡βb2=0\frac{{\cos \alpha \,\,\cos \beta }}{{{a^2}}} + \frac{{\sin \alpha \,\,\sin \beta }}{{{b^2}}} = 0, then
View answer and explanation
Correct option: A
tan⁡α  tan⁡β=b2(x2−a2)a2(y2−b2)   and   x2+y2=a2−b2\tan \alpha \,\,\tan \beta = \frac{{{b^2}\left( {{x^2} - {a^2}} \right)}}{{{a^2}\left( {{y^2} - {b^2}} \right)}}\;\,\mathrm{and}\,\;{x^2} + {y^2} = {a^2} - {b^2}
Option A is correct because it derives the relationship between tan α tan β, x, y, a, and b based on the given equations and trigonometric identities.
81 Easy
81 of 394
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.
82 Easy
82 of 394
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.
83 Easy
83 of 394
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.
84 Medium
84 of 394
Application +4 | -1
If π<α<3π2\pi < \alpha < \frac{3 \pi}{2}, then 1−cos⁡α1+cos⁡α+1+cos⁡α1−cos⁡α=\sqrt{\frac{1-\cos \alpha}{1+\cos \alpha}}+\sqrt{\frac{1+\cos \alpha}{1-\cos \alpha}}=
View answer and explanation
Correct option: B
−2sin⁡α-\frac{2}{\sin \alpha}
Option B (−2sin⁡α-\frac{{2}}{{\sin \alpha}}) is correct because, within the given range for alpha, the expression simplifies to this value considering the sign of sine in that quadrant.
85 Easy
85 of 394
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.
86 Easy
86 of 394
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.
87 Hard
87 of 394
Application +4 | -1
The minimum value of the function f(x)=3tan⁡x+4cot⁡xf(x)=3 \tan x+4 \cot x where x∈(0,π2)x \in\left(0, \frac{\pi}{2}\right) is
View answer and explanation
Correct option: B
2122 \sqrt{12}
Option B (2 sqrt(12)) is correct because the minimum value of the function can be found using AM >= GM inequality, which gives 3tan(x) + 4cot(x) >= 2*sqrt(3tan(x)*4cot(x)) = 2*sqrt(12).
88 Easy
88 of 394
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.
89 Medium
89 of 394
Analyzing +4 | -1
If y=tan⁡  xtan⁡  3xy = \frac{{\tan \,\,x}}{{\tan \,\,3x}}, then
View answer and explanation
Correct option: B
y  ∉  [1/3,   3]y\,\, \notin \,\,\left[ {1/3,\;\,3} \right]
Option B (y∉[1/3,3]y \notin [1/3, 3]) is correct because solving for y=tan⁡xtan⁡3xy = \frac{{\tan x}}{{\tan 3x}} leads to the conclusion that y cannot lie within the closed interval [1/3, 3].
90 Medium
90 of 394
Application +4 | -1
If cos⁡α+cos⁡β=a,sin⁡α+sin⁡β=b\cos \alpha+\cos \beta=a, \sin \alpha+\sin \beta=b and α−β=2θ\alpha-\beta=2 \theta, then cos⁡3θcos⁡θ=\frac{\cos 3 \theta}{\cos \theta}=
View answer and explanation
Correct option: B
a2+b2−3a^2+b^2-3
Option B (a2+b2−3a^2+b^2-3) is correct because by using the given relations and trigonometric identities, the expression cos⁡3θcos⁡θ\frac{\cos 3 \theta}{\cos \theta} simplifies to a2+b2−3a^2+b^2-3.
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