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JEE • MATHEMATICS • TRIGONOMETRY RATIOS

Trigonometry ratios practice · Easy

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121 Easy
121 of 218
Understanding +4 | -1
If sin⁡  θ1+sin⁡  θ2+sin⁡  θ3=3\sin \,\,{\theta _1} + \sin \,\,{\theta _2} + \sin \,\,{\theta _3} = 3, then cos⁡  θ1+cos⁡  θ2+cos⁡  θ3=\cos \,\,{\theta _1} + \cos \,\,{\theta _2} + \cos \,\,{\theta _3} =
View answer and explanation
Correct option: D
0\mathrm{0}
Option D (0) is correct because the maximum value of sin⁡θ\sin \theta is 1, which implies θ1=θ2=θ3=π2\theta_1 = \theta_2 = \theta_3 = \frac{\pi}{2}, making the sum of cosines zero.
122 Easy
122 of 218
Analyzing +4 | -1
If a  sin⁡2  x+b  cos⁡2  x=c,    b  sin⁡2  y+a  cos⁡2  y=da\,\,{\sin ^2}\,\,x + b\,\,{\cos ^2}\,\,x = c,\,\,\;b\,\,{\sin ^2}\,\,y + a\,\,{\cos ^2}\,\,y = d and a  tan⁡  x=b  tan⁡  ya\,\,\tan \,\,x = b\,\,\tan \,\,y, then a2b2\frac{{{a^2}}}{{{b^2}}} is equal to

View answer and explanation
Correct option: B
(a−d)  (c−a)(b−c)  (d−b)\frac{{\left( {a - d} \right)\,\,\left( {c - a} \right)}}{{\left( {b - c} \right)\,\,\left( {d - b} \right)}}
Option B ((a−d)(c−a)(b−c)(d−b)\frac{{\left( {a - d} \right)\left( {c - a} \right)}}{{\left( {b - c} \right)\left( {d - b} \right)}}) is correct because it is derived from manipulating the given equations and using trigonometric identities.
123 Easy
123 of 218
Application +4 | -1
If cos⁡θ=cos⁡αcos⁡β\cos \theta=\cos \alpha \cos \beta, then tan⁡θ+α2tan⁡θ−α2\tan \frac{\theta+\alpha}{2} \tan \frac{\theta-\alpha}{2} is equal to
View answer and explanation
Correct option: D
cot⁡2(β/2)\cot ^2(\beta / 2)
Option B (tan⁡2(β/2)\tan ^2(\beta / 2)) is correct because the given trigonometric expression simplifies to tan⁡2(β/2)\tan^2(\beta/2) using trigonometric identities.
124 Easy
124 of 218
Application +4 | -1
If Sn=cos⁡n  θ+sin⁡n  θ  {S_n} = {\cos ^n}\,\,\theta + {\sin ^n}\,\,\theta \;, then the value of 3S4−2S63{S_4} - 2{S_6} is given by
View answer and explanation
Correct option: C
1
Option C (1) is correct because substituting the given expression and simplifying using trigonometric identities results in 1.
125 Easy
125 of 218
Application +4 | -1
The function f(x)=acos⁡x+btan⁡x+xf\left( x \right) = a\cos x + b\tan x + x has extreme values at x=0x = 0 and x=π6x = \frac{\pi }{6}, then
View answer and explanation
Correct option: A
a=−23,b=−1a = - \frac{2}{3},b = - 1
Option C (0,0) is correct because when the tangent is perpendicular to the x-axis, the slope is infinite, implying t=0, which gives the point (0,0).
126 Easy
126 of 218
Application +4 | -1
If cos⁡θ+sin⁡ϕ=m\cos \theta+\sin \phi=m and sin⁡θ+cos⁡ϕ=n\sin \theta+\cos \phi=n, then 12(m2+n2−2)\frac{1}{2}\left(m^2+n^2-2\right) is equal to
View answer and explanation
Correct option: B
sin⁡(θ+ϕ)\sin (\theta+\phi)
Option B (sin⁡(θ+ϕ)\sin (\theta+\phi)) is correct because simplifying the expression 12(m2+n2−2)\frac{{1}}{{2}}(m^2 + n^2 - 2) using the given equations results in sin⁡(θ+ϕ)\sin (\theta + \phi).
127 Easy
127 of 218
Analyzing +4 | -1
If sin⁡4α+4cos⁡4 β+2=42  sin⁡α  cos⁡β ;   α,β  ∈  [0, π],{\sin ^4}\alpha + 4{\cos ^4}\,\beta + 2 = 4\sqrt 2 \,\,\sin \alpha \,\,\cos \beta \,;\,\,\,\alpha ,\beta \,\, \in \,\,\left[ {0,\,\pi } \right], then cos⁡(α+β)−cos⁡(α−β)\cos \left( {\alpha + \beta } \right) - \cos \left( {\alpha - \beta } \right) is equal to
View answer and explanation
Correct option: C
−2- \sqrt 2
Option C (−2- \sqrt 2) is correct because by applying AM >= GM inequality, we find the values of sin⁡α\sin \alpha and cos⁡β\cos \beta, and then calculate the expression cos⁡(α+β)−cos⁡(α−β)\cos (\alpha + \beta ) - \cos (\alpha - \beta ).
128 Easy
128 of 218
Application +4 | -1
If sin⁡x+cos⁡y=a\sin x+\cos y=a and cos⁡x+sin⁡y=b\cos x+\sin y=b, then tan⁡x−y2\tan \frac{x-y}{2} is equal to
View answer and explanation
Correct option: D
a−ba+b\frac{a-b}{a+b}
Option D (a−ba+b\frac{{a-b}}{{a+b}}) is correct because it is derived by using sum-to-product trigonometric identities and then applying the tangent half-angle formula.
129 Easy
129 of 218
Application +4 | -1
The value of cos⁡2π7+cos⁡4π7+\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+ cos⁡6π7+cos⁡7π7\cos \frac{6 \pi}{7}+\cos \frac{7 \pi}{7} is
View answer and explanation
Correct option: D
−3/2-3 / 2
Option D (-3/2) is correct because by using the sum-to-product trigonometric identities, the expression simplifies to -3/2.
130 Easy
130 of 218
Application +4 | -1
With usual notations in a △ABC\triangle A B C, b2−c2asec⁡A+c2−a2bsec⁡B+a2−b2csec⁡C\frac{b^2-c^2}{a \sec A}+\frac{c^2-a^2}{b \sec B}+\frac{a^2-b^2}{c \sec C} is equal to
View answer and explanation
Correct option: B
0
Option B (0) is correct because by applying the Law of Cosines and simplifying, the expression evaluates to zero.
131 Easy
131 of 218
Application +4 | -1
sin⁡2  5∘+sin⁡2  10∘+sin⁡2  15∘+...+sin⁡2  90∘{\sin ^2}\,\,5^\circ + {\sin ^2}\,\,10^\circ + {\sin ^2}\,\,15^\circ + ... + {\sin ^2}\,\,90^\circ is equal to
View answer and explanation
Correct option: C
9  129\,\,\frac{1}{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.
132 Easy
132 of 218
Application +4 | -1
If π2<θ<π\frac{\pi }{2} < \theta < \pi, then 1−sin⁡  θ1+sin⁡  θ+1+sin⁡  θ1−sin⁡  θ\sqrt {\frac{{1 - \sin \,\,\theta }}{{1 + \sin \,\,\theta }}} + \sqrt {\frac{{1 + \sin \,\,\theta }}{{1 - \sin \,\,\theta }}} is equal to
View answer and explanation
Correct option: B
− 2  sec⁡  θ- \,2\,\,\sec \,\,\theta
Option B ( − 2sec⁡θ\ - \,2\sec \theta) is correct because the expression simplifies to −2sec⁡θ-2\sec \theta when π2<θ<π\frac{\pi}{2} < \theta < \pi, considering the sign of cos⁡θ\cos \theta in the second quadrant.
133 Easy
133 of 218
Understanding +4 | -1
If sin⁡θ=−12\sin \theta=-\frac{1}{\sqrt{2}} and tan⁡θ=1\tan \theta=1, then θ\theta lies in which quadrant
View answer and explanation
Correct option: C
Third
Option C (Third) is correct because sine is negative and tangent is positive only in the third quadrant.
134 Easy
134 of 218
Application +4 | -1
At what time between 2 and 3 ' OO clock, both the hands will be at right angle?
View answer and explanation
Correct option: A
1641116 \frac{4}{11} minutes past 2
Option A (1641116 \frac{4}{11} minutes past 2) is correct because it is derived from the relative speed concept of the hour and minute hands and solving for the time when the angle between them is 90 degrees.
135 Easy
135 of 218
Application +4 | -1
If 0<θ<π/20 < \theta<\pi / 2, and f(θ)=(1+cot⁡θ−cosec⁡θ)f(\theta)=(1+\cot \theta-\operatorname{cosec} \theta) (1+tan⁡θ+sec⁡θ)(1+\tan \theta+\sec \theta), then f(π/4)f(\pi / 4) is equal to
View answer and explanation
Correct option: C
2
Option C (2) is correct because the expression simplifies to a constant value of 2 after applying trigonometric identities.

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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
136 Easy
136 of 218
Application +4 | -1
If cos⁡  x+cos⁡2x=1\cos \,\,x + {\cos ^2}x = 1, then the value of sin⁡12x+3  sin⁡10x+3  sin⁡8x+sin⁡6x−1{\sin ^{12}}x + 3\,\,{\sin ^{10}}x + 3\,\,{\sin ^8}x + {\sin ^6}x - 1, is equal to
View answer and explanation
Correct option: D
0\mathrm{0}
Option D (0) is correct because the expression simplifies to 0 after substituting cos x = sin²x.
137 Easy
137 of 218
Application +4 | -1
If cosec⁡θ=p+qp−q\operatorname{cosec} \theta=\frac{p+q}{p-q} where p>q>0p>q>0, then ∣cot⁡(π4+θ2)∣\left|\cot \left(\frac{\pi}{4}+\frac{\theta}{2}\right)\right| is equal to:
View answer and explanation
Correct option: A
pq\sqrt{\frac{p}{q}}
The correct answer is A (pq\sqrt{\frac{p}{q}}) because by using trigonometric identities and the given information, we can simplify the expression to obtain the result.
138 Easy
138 of 218
Application +4 | -1
The value of 3+cot⁡76∘cot⁡16∘cot⁡76∘+cot⁡16∘\frac{3+\cot 76^{\circ} \cot 16^{\circ}}{\cot 76^{\circ}+\cot 16^{\circ}} is
View answer and explanation
Correct option: D
cot⁡44∘\cot 44^{\circ}
Option D (cot⁡44∘\cot 44^{\circ}) is correct because the expression simplifies to tan⁡46∘\tan 46^{\circ}, which is equal to cot⁡44∘\cot 44^{\circ}, using trigonometric identities.
139 Easy
139 of 218
Application +4 | -1
The value of cos⁡(270∘+θ)⋅cos⁡(90∘−θ)−\cos \left(270^{\circ}+\theta\right) \cdot \cos \left(90^{\circ}-\theta\right)- sin⁡(270∘−θ)cos⁡θ\sin \left(270^{\circ}-\theta\right) \cos \theta is
View answer and explanation
Correct option: D
1
Option D (1) is correct because after applying trigonometric identities and simplifying, the expression reduces to sin⁡2θ+cos⁡2θ\sin^2 \theta + \cos^2 \theta, which equals 1.
140 Easy
140 of 218
Application +4 | -1
cos⁡2(π6+θ)−sin⁡2(π6−θ)=\cos ^2\left(\frac{\pi}{6}+\theta\right)-\sin ^2\left(\frac{\pi}{6}-\theta\right)=
View answer and explanation
Correct option: A
12cos⁡2θ\frac{1}{2} \cos 2 \theta
Option A 12cos⁡2θ\frac{{1}}{{2}} \cos 2 \theta is correct because it is obtained by applying the trigonometric identity for the difference of squares of cosines and sines.
141 Easy
141 of 218
Understanding +4 | -1
In a cyclic quadrilateral ABCDABCD, he value of cos⁡  A+cos⁡  B+cos⁡  C+cos⁡  D\cos \,\,A + \cos \,\,B + \cos \,\,C + \cos \,\,D, is
View answer and explanation
Correct option: B
0\mathrm{0}
Option B (0) is correct because in a cyclic quadrilateral, opposite angles are supplementary (A + C = 180° and B + D = 180°), leading to cos A = -cos C and cos B = -cos D, thus their sum is zero.
142 Easy
142 of 218
Application +4 | -1
The value of cos⁡255∘+sin⁡195∘\cos 255^{\circ}+\sin 195^{\circ} is
View answer and explanation
Correct option: C
−(3−12)-\left(\frac{\sqrt{3}-1}{\sqrt{2}}\right)
Option C (-(3−12\frac{\sqrt{3}-1}{\sqrt{2}})) is correct because it is the simplified value of the given trigonometric expression.
143 Easy
143 of 218
Application +4 | -1
The value of
sin⁡2  5∘+sin⁡2  10∘+sin⁡2  15∘+…  +sin⁡2  85∘+sin⁡2  90∘{\sin ^2}\,\,5^\circ + {\sin ^2}\,\,10^\circ + {\sin ^2}\,\,15^\circ + \ldots \,\, + {\sin ^2}\,\,85^\circ + {\sin ^2}\,\,90^\circ, is
View answer and explanation
Correct option: C
9  129\,\,\frac{1}{2}
Option C (9 129\,\frac{1}{2}) is the correct answer because by pairing terms and using trigonometric identities, the sum simplifies to this value.
144 Easy
144 of 218
Application +4 | -1
Value of sin⁡πn+sin⁡3πn+sin⁡5πn+……\sin \frac{\pi}{n}+\sin \frac{3 \pi}{n}+\sin \frac{5 \pi}{n}+\ldots \ldots to nn terms is:
View answer and explanation
Correct option: B
0
Option B (0) is correct because the given series is a sum of sines in an arithmetic progression, and using the formula for the sum, the expression simplifies to 0.
145 Easy
145 of 218
Application +4 | -1
sin⁡2(π8+A2)−sin⁡2(π8−A2)\sin ^2\left(\frac{\pi}{8}+\frac{A}{2}\right)-\sin ^2\left(\frac{\pi}{8}-\frac{A}{2}\right) is equal to:
View answer and explanation
Correct option: A
(1/2)sin⁡A(1 / \sqrt{2}) \sin A
Option A (12sin⁡A\frac{1}{\sqrt{2}} \sin A) is correct because using the identity sin⁡2A−sin⁡2B=sin⁡(A−B)sin⁡(A+B)\sin^2 A - \sin^2 B = \sin(A-B)\sin(A+B), the expression simplifies to 12sin⁡A\frac{1}{\sqrt{2}} \sin A.
146 Easy
146 of 218
Application +4 | -1
Value of cos⁡(3π2+x)cos⁡(2π+x)[cot⁡(3π2−x)+cot⁡(2π+x)]\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right] is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the given expression simplifies to 1 using trigonometric identities.
147 Easy
147 of 218
Application +4 | -1
If tan⁡  θ=x−14x\tan \,\,\theta = x - \frac{1}{{4x}}, then sec⁡  θ−tan⁡  θ\sec \,\,\theta - \tan \,\,\theta is equal to
View answer and explanation
Correct option: A
− 2x,  12x- \,2x,\,\,\frac{1}{{2x}}
Option A ( − 2x, 12x\ - \,2x,\,\frac{{1}}{{2x}}) is correct because it is derived by using trigonometric identities and solving for sec⁡ θ−tan⁡ θ\sec \,\theta - \tan \,\theta.
148 Easy
148 of 218
Application +4 | -1
The maximum and minimum value of f(x)f(x) =absin⁡x+b1−a2cos⁡x+c=a b \sin x+b \sqrt{1-a^2} \cos x+c lie in the interval ( assuming ∣a∣<1,b>0)|a|<1, b>0)
View answer and explanation
Correct option: C
[c−b,b+c][c-b, b+c]
Option C ([c-b, b+c]) is correct because the maximum and minimum values of the function lie in the interval [c-b, b+c], which is found by rewriting the function as a sine function and using its amplitude.
149 Easy
149 of 218
Application +4 | -1
If 0<x<π0 < x < \pi, and sin⁡x+cos⁡x=1/3\sin x+\cos x=1 / 3, then a possible value of tan⁡x\tan x is
View answer and explanation
Correct option: A
18(−9+17)\frac{1}{8}(-9+\sqrt{17})
The correct answer is A (18(−9+17)\frac{1}{8}(-9+\sqrt{17})) because by squaring the given equation and using trigonometric identities, we can solve for tan x.
150 Easy
150 of 218
Application +4 | -1
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 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), the expression simplifies to 5+18\frac{{\sqrt{5}+1}}{{8}}.
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