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JEE • MATHEMATICS • MULTIPLE AND SUBMULTIPLE ANGLES

Multiple and submultiple angles 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 1-30 of 66 questions
1 Easy
1 of 66
Application +4 | -1
If 1+cos⁡  x=k1 + \cos \,\,x = k, where xx is acute, then sin⁡  x2\sin \,\,\frac{x}{2} is
View answer and explanation
Correct option: D
2−k2\sqrt {\frac{{2 - k}}{2}}
Option D (2−k2\sqrt {\frac{{2 - k}}{2}}) is correct because by using the half-angle formula for sine and the given equation, we can derive this expression.
2 Easy
2 of 66
Application +4 | -1
If tan⁡  α=ba,   a>b>0\tan \,\,\alpha = \frac{b}{a},\,\;a > b > 0 and if 0<α<π40 < \alpha < \frac{\pi }{4}, 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: A
2  sin⁡  αcos⁡  2α\frac{{2\,\,\sin \,\,\alpha }}{{\sqrt {\cos \,\,2\alpha } }}
The correct answer is A (2sin⁡αcos⁡,2α\frac{{2\sin \alpha }}{{\sqrt {\cos \\,2\alpha } }}) because simplifying the given expression using trigonometric identities leads to this result.
3 Easy
3 of 66
Application +4 | -1
The value of

cos⁡π65cos⁡2π65cos⁡4π65cos⁡8π65cos⁡16π65cos⁡32π65\cos \frac{\pi}{65} \cos \frac{2 \pi}{65} \cos \frac{4 \pi}{65} \cos \frac{8 \pi}{65} \cos \frac{16 \pi}{65} \cos \frac{32 \pi}{65}
is equal to
View answer and explanation
Correct option: A
164\frac{1}{64}
Option A (164\frac{1}{64}) is correct because the given expression can be simplified using the double angle formula repeatedly.
4 Easy
4 of 66
Application +4 | -1
The value of   cos⁡  π65  cos⁡  2 π65  cos  4 π65  …  cos⁡  32 π65\mathrm{\;}\cos \,\,\frac{\mathrm{\pi }}{{65}}\,\,\cos \,\,\frac{{2\,\mathrm{\pi }}}{{65}}\,\,\mathrm{cos}\,\,\frac{{4\,\mathrm{\pi }}}{{65}}\,\, \ldots \,\,\cos \,\,\frac{{32\,\mathrm{\pi }}}{{65}} is
View answer and explanation
Correct option: B
164\frac{1}{{64}}
Option B (164\frac{1}{{64}}) is correct because repeated application of the identity sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2 \sin x \cos x simplifies the expression.
5 Easy
5 of 66
Analyzing +4 | -1
If   α,  β,  γ, δ\;\alpha ,\,\,\beta ,\,\,\gamma ,\,\delta are four solutions of the equation tan⁡  (θ+π4)=3  tan⁡ 3θ\tan \,\,\left( {\mathrm{\theta } + \frac{\pi }{4}} \right) = 3\,\,\tan \,3\mathrm{\theta }, then
tan⁡  α  tan⁡  β  tan⁡  γ  tan⁡  δ  \tan \,\,\alpha \,\,\tan \,\,\beta \,\,\tan \,\,\gamma \,\,\tan \,\,\delta \; equals
View answer and explanation
Correct option: C
−13- \frac{1}{3}
Option C (−13\frac{{ - 1}}{3}) is correct because the product of the roots of the quartic equation formed from the given trigonometric equation is equal to −13\frac{{ - 1}}{3}.
6 Easy
6 of 66
Application +4 | -1
Simplest form of 22+2+2+2  cos⁡  4x\frac{2}{{\sqrt {2 + \sqrt {2 + \sqrt {2 + 2\,\,\cos \,\,4x} } } }} is
View answer and explanation
Correct option: A
sec⁡  x2\sec \,\,\frac{x}{2}
Option A (sec⁡  x2\sec \,\,\frac{x}{2}) is correct because it is obtained by repeatedly using the identity 2cos⁡2θ=1+cos⁡2θ2\cos^2 \theta = 1 + \cos 2\theta and simplifying the expression.
7 Easy
7 of 66
Application +4 | -1
If tan⁡2θ=2tan⁡2ϕ+1,{\tan ^2}\theta = 2{\tan ^2}\phi + 1, then cos⁡2θ+sin⁡2ϕ\cos 2\theta + {\sin ^2}\phi is equal to
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the expression cos⁡2θ+sin⁡2ϕ\cos 2\theta + \sin^2 \phi simplifies to 0, which is not among the given options.
8 Easy
8 of 66
Application +4 | -1
tan⁡6  π9−33  tan⁡4  π9+27  tan⁡2  π9={\tan ^6}\,\,\frac{\pi }{9} - 33\,\,{\tan ^4}\,\,\frac{\pi }{9} + 27\,\,{\tan ^2}\,\,\frac{\pi }{9} =
View answer and explanation
Correct option: C
3
Option C (3) is correct because by using the triple angle formula for tangent, the expression simplifies to 3.
9 Easy
9 of 66
Application +4 | -1
If   tan  α/2   and tan⁡β/2    \mathrm{\;tan}\,\,\alpha /2\,\;\mathrm{and}\,\tan \beta /2\;\; are the roots of the equation 8 x2−  26 x+15=08\,{x^2} - \;26\,x + 15 = 0 then
cos⁡ (α+β)\cos \,\left( {\alpha + \beta } \right) is equal to
View answer and explanation
Correct option: A
−627725- \frac{{627}}{{725}}
Option A is correct because by finding the sum and product of the roots of the quadratic equation, we can determine \tan \left( {\frac{{\alpha + \mathrm{\beta }}}}{2} \right)\tan \left( {\frac{{\alpha + \mathrm{\beta }}}}{2} \right) and then calculate cos⁡(α+β)\cos \left( {\alpha + \mathrm{\beta }} \right) using trigonometric identities.
10 Easy
10 of 66
Application +4 | -1
The value of cot⁡x−tan⁡xcot⁡  2x\frac{{\cot x - \tan x}}{{\cot \,\,2x}} is
View answer and explanation
Correct option: B
2
Option B (2) is correct because simplifying the given trigonometric expression using identities leads to the value 2.
11 Easy
11 of 66
Application +4 | -1
1−tan⁡2  (45∘−A)1+tan⁡2  (45∘−A)\frac{{1 - {{\tan }^2}\,\,\left( {45^\circ - A} \right)}}{{1 + {{\tan }^2}\,\,\left( {45^\circ - A} \right)}} is equal to
View answer and explanation
Correct option: A
sin⁡  2A\sin \,\,2A
Option B (ab\frac{a}{b}) is correct because by applying the componendo and dividendo rule and trigonometric identities, the expression simplifies to ab\frac{a}{b}.
12 Easy
12 of 66
Application +4 | -1
If x+1x=2  cos⁡  θx + \frac{1}{x} = 2\,\,\cos \,\,\theta, then xn+1xn{x^n} + \frac{1}{{{x^n}}} is equal to
View answer and explanation
Correct option: B
2  cos⁡  nθ2\,\,\cos \,\,n\theta
Option B (2 cos⁡ nθ2\,\cos \,n\theta) is correct because if x+1x=2 cos⁡θx + \frac{1}{x} = 2\,\cos \theta, then xn+1xn=2 cos⁡ nθ{x^n} + \frac{1}{{{x^n}}} = 2\,\cos \,n\theta.
13 Easy
13 of 66
Application +4 | -1
The value of (3+cot⁡  76∘  cot⁡  16∘)cot⁡  76∘  +cot⁡  16∘\frac{{\left( {3 + \cot \,\,76^\circ \,\,\cot \,\,16^\circ } \right)}}{{\cot \,\,76^\circ \,\, + \cot \,\,16^\circ }} is
View answer and explanation
Correct option: A
cot⁡  44∘\cot \,\,44^\circ
Option A (cot⁡,44∘\cot \\,44^\circ) is correct because simplifying the expression using trigonometric identities leads to tan⁡46∘\tan 46^\circ, which is equal to cot⁡44∘\cot 44^\circ.
14 Easy
14 of 66
Analyzing +4 | -1
The value of cos⁡3(π8)⋅cos⁡(3π8)+sin⁡3(π8)⋅sin⁡(3π8){\cos ^3}\left( {\frac{\pi }{8}} \right) \cdot \cos \left( {\frac{{3\pi }}{8}} \right) + {\sin ^3}\left( {\frac{\pi }{8}} \right) \cdot \sin \left( {\frac{{3\pi }}{8}} \right) is
View answer and explanation
Correct option: B
122\frac{1}{{2\sqrt 2 }}
Option B (122\frac{1}{{2\sqrt 2 }}) is correct because it is the simplified value of the given trigonometric expression after applying trigonometric identities.
15 Easy
15 of 66
Application +4 | -1
If sinθ  =−45  \mathrm{sin \theta \;} = - \frac{4}{{5\;}} and θ  \mathrm{\theta \;} lies in the third quadrant, then cos  θ2  \mathrm{cos}\,\,\frac{\mathrm{\theta }}{{2\;}} is equal to
View answer and explanation
Correct option: B
−15- \frac{1}{{\sqrt 5 }}
Option B (-1/√5) is correct because, given sin θ is negative and θ is in the third quadrant, cos(θ/2) will be negative, and the calculation yields -1/√5.

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16 Easy
16 of 66
Application +4 | -1
If α\alpha is an acute angle and sin⁡  α2=x−12x\sin \,\,\frac{\alpha }{2} = \sqrt {\frac{{x - 1}}{{2x}}}, then tan⁡  α\tan \,\,\alpha is
View answer and explanation
Correct option: C
x2−1\sqrt {{x^2} - 1}
Option C (x2−1\sqrt {{x^2} - 1}) is the correct answer because it is derived from trigonometric identities and simplification of the given equation.
17 Easy
17 of 66
Application +4 | -1
If x=tan⁡15∘,  y=cosec   75∘x = \tan 15^\circ ,\;y = \mathrm{cosec\;}\,75^\circ and z=4sin⁡18∘z = 4\sin 18^\circ, then

View answer and explanation
Correct option: A
x<y<zx < y < z
Option D (x<z<yx < z < y) is correct because after calculating the values of x, y, and z, we find that tan⁡15∘<4sin⁡18∘<csc⁡75∘\tan 15^\circ < 4\sin 18^\circ < \csc 75^\circ.
18 Easy
18 of 66
Analyzing +4 | -1
If α\alpha and β\beta are acute angles cos⁡  2α=3cos⁡  2β−13−cos⁡  2β\cos \,\,2\alpha = \frac{{3\cos \,\,2\beta - 1}}{{3 - \cos \,\,2\beta }}, then tan⁡α  cot⁡β=\tan \alpha \,\,\cot \beta =
View answer and explanation
Correct option: B
2\sqrt 2
Option B (2\sqrt 2) is correct because by applying componendo and dividendo and simplifying, we find that tan⁡αcot⁡β=2\tan \alpha \cot \beta = \sqrt 2.
19 Easy
19 of 66
Analyzing +4 | -1
If tan⁡  x2=cosec    x−sin⁡  x\tan \,\,\frac{x}{2} = \mathrm{cosec}\;\,\,x - \sin \,\,x, then the value of tan⁡2  x2{\tan ^2}\,\,\frac{x}{2}, is
View answer and explanation
Correct option: D
− 2+5- \,2 + \sqrt 5
Option D (−2+5-2 + \sqrt{5}) is correct because solving the trigonometric equation for tan⁡2x2\tan^2 \frac{x}{2} leads to the value −2+5-2 + \sqrt{5}.
20 Easy
20 of 66
Application +4 | -1
The value of cos⁡2  76∘+cos⁡2  16−cos⁡76∘  cos⁡  16∘{\cos ^2}\,\,76^\circ + {\cos ^2}\,\,16 - \cos 76^\circ \,\,\cos \,\,16^\circ, is
View answer and explanation
Correct option: D
3/43/4
Option B (1) is correct because by using the cosine rule and the given condition C = 60°, we can simplify the expression to 1.
21 Easy
21 of 66
Application +4 | -1
Let α, β\alpha ,\,\beta be such that π<α−β<3π\pi < \alpha - \beta < 3\pi. If sin⁡  α+sin⁡β=−2165\sin \,\,\alpha + \sin \beta = - \frac{{21}}{{65}} and cos⁡  α+cos⁡  β=−2765\cos \,\,\alpha + \cos \,\,\beta = - \frac{{27}}{{65}}, then the value of cos⁡  (α−β)2\cos \,\,\frac{{\left( {\alpha - \beta } \right)}}{2} is
View answer and explanation
Correct option: A
−3130- \frac{3}{{\sqrt {130} }}
Option A (−3130- \frac{3}{{\sqrt {130} }}) is correct because, after squaring and adding the given equations and using trigonometric identities, we find the value of cos⁡(α−β)2\cos \frac{{\left( {\alpha - \beta } \right)}}{2} considering the given range of α−β\alpha - \beta.
22 Easy
22 of 66
Application +4 | -1
cos⁡θcos⁡2θcos⁡4θ…cos⁡2n−1θ\cos \theta \cos 2 \theta \cos 4 \theta \ldots \cos 2^{n-1} \theta is equal to
View answer and explanation
Correct option: A
sin⁡2nθ2nsin⁡θ\frac{\sin 2^n \theta}{2^n \sin \theta}
Option A is correct because the given expression can be simplified to sin⁡2nθ2nsin⁡θ\frac{{\sin 2^n \theta}}{{2^n \sin \theta}} using trigonometric identities.
23 Easy
23 of 66
Analyzing +4 | -1
If cos⁡  2B=cos⁡  (A+C)cos⁡  (A−C)  \cos \,\,2B = \frac{{\cos \,\,\left( {A + C} \right)}}{{\cos \,\,\left( {A - C} \right)}}\;, then tan⁡  A,  tan⁡  B,  tan⁡  C\tan \,\,A,\,\,\tan \,\,B,\,\,\tan \,\,C are in
View answer and explanation
Correct option: B
GP
The correct answer is B (GP) because after simplification and using trigonometric identities, it's shown that tan⁡2B=tan⁡Atan⁡C\tan^2 B = \tan A \tan C, which is the condition for GP.
24 Easy
24 of 66
Application +4 | -1
If x=log⁡  [cot⁡(π4+θ)]x = \log \,\,\left[ {\cot \left( {\frac{\pi }{4} + \theta } \right)} \right], then the value of sinh⁡  x\sinh \,\,x is
View answer and explanation
Correct option: B
−tan⁡  2θ- \tan \,\,2\theta
Option B (−tan⁡ 2θ- \tan \,2\theta) is correct because simplifying the expression sinh⁡x=ex−e−x2\sinh x = \frac{{{e^x} - {e^{ - x}}}}{2} with the given value of xx yields −tan⁡ 2θ- \tan \,2\theta.
25 Easy
25 of 66
Application +4 | -1
The value of cot⁡2  θ+1cot⁡2  θ−1\frac{{{{\cot }^2}\,\,\mathrm{\theta } + 1}}{{{{\cot }^2}\,\,\mathrm{\theta } - 1}} is equal to
View answer and explanation
Correct option: D
sec⁡  2θ\sec \,\,2\mathrm{\theta }
Option D (sec⁡ 2θ\sec \,2\mathrm{\theta }) is correct because the given expression can be simplified using trigonometric identities to equal sec⁡ 2θ\sec \,2\mathrm{\theta }.
26 Easy
26 of 66
Analyzing +4 | -1
If sin⁡θ=3sin⁡(θ+2α)\sin \theta=3 \sin (\theta+2 \alpha), then the value of tan⁡(θ+α)+2tan⁡α\tan (\theta+\alpha)+2 \tan \alpha is
View answer and explanation
Correct option: D
0
Option D (0) is correct because simplifying the given equation leads to tan⁡(θ+α)+2tan⁡α=0\tan (\theta + \alpha) + 2 \tan \alpha = 0.
27 Easy
27 of 66
Application +4 | -1
The value of cot⁡π24\cot \frac{\pi }{{24}} is
View answer and explanation
Correct option: B
2+3+2+6\sqrt 2 + \sqrt 3 + 2 + \sqrt 6
Option B (2+3+2+6\sqrt 2 + \sqrt 3 + 2 + \sqrt 6) is correct because it is the simplified value of cot(π/24) after applying trigonometric identities.
28 Easy
28 of 66
Analyzing +4 | -1
tan⁡  9∘−tan⁡  27∘−tan⁡  63∘+tan⁡  81∘\tan \,\,9^\circ - \tan \,\,27^\circ - \tan \,\,63^\circ + \tan \,\,81^\circ is equal to
View answer and explanation
Correct option: D
4
Option D (4) is correct because simplifying the trigonometric expression results in a value of 4.
29 Easy
29 of 66
Application +4 | -1
tan⁡  6∘  tan⁡  42∘  tan⁡  66∘  tan⁡  78∘  \tan \,\,6^\circ \,\,\tan \,\,42^\circ \,\,\tan \,\,66^\circ \,\,\tan \,\,78^\circ \mathrm{\;} is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because the expression simplifies to 1 using trigonometric identities.
30 Easy
30 of 66
Application +4 | -1
The value of cos⁡  9∘−sin⁡  9∘\cos \,\,9^\circ - \sin \,\,9^\circ is
View answer and explanation
Correct option: B
5−52\frac{{\sqrt {5 - \sqrt 5 } }}{2}
Option B (5−52\frac{{\sqrt {5 - \sqrt 5 } }}{2}) is correct because it is derived from simplifying the expression 1−sin⁡18∘\sqrt{1 - \sin 18^\circ} using the known value of sin⁡18∘\sin 18^\circ.
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