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JEE • MATHEMATICS • MULTIPLE AND SUBMULTIPLE ANGLES

Multiple and submultiple angles practice · Mixed difficulty

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Showing 31-60 of 122 questions
31 Easy
31 of 122
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.
32 Easy
32 of 122
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.
33 Easy
33 of 122
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.
34 Easy
34 of 122
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.
35 Easy
35 of 122
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.
36 Easy
36 of 122
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.
37 Easy
37 of 122
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}.
38 Easy
38 of 122
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.
39 Medium
39 of 122
Application +4 | -1
If 0<A<π60 < A < \frac{\pi }{6} and sin⁡A+cos⁡A=72\sin A + \cos A = \frac{{\sqrt 7 }}{2}, then tan⁡A2=\tan \frac{A}{2} =
View answer and explanation
Correct option: A
7−23\frac{{\sqrt 7 - 2}}{3}
Option A (7−23\frac{{\sqrt 7 - 2}}{3}) is correct because it is the solution for tan⁡A2\tan \frac{A}{2} derived from the given trigonometric equation and the constraint on A.
40 Easy
40 of 122
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.
41 Easy
41 of 122
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.
42 Easy
42 of 122
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.
43 Medium
43 of 122
Application +4 | -1
If 0<θ<π0 < \theta < \pi, then 2+  2  +2  +…+2+2  cos⁡  θ\sqrt {2 + \,\,\sqrt {2\,\, + \sqrt {2\,\, + \ldots + \sqrt {2 + 2\,\,\cos \,\,\theta } } } } there being n number of 2′ s\mathrm{2'}\,\mathrm{s} is equal to
View answer and explanation
Correct option: A
2  cos⁡  θ2n2\,\,\cos \,\,\frac{\theta }{{{2^n}}}
Option A (2cos⁡θ2n2\cos \frac{\theta}{{{{2^n}}}}) is correct because it represents the simplified form of the nested square root expression after applying the identity 1+cos⁡θ=2cos⁡2θ21 + \cos \theta = 2\cos^2 \frac{\theta}{2} n times.
44 Easy
44 of 122
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.
45 Hard
45 of 122
Analyzing +4 | -1
For x∈Rx \in R,

tan⁡  x+12  tan⁡  x2  +  122  tan⁡  x22+…+12n−1  tan⁡  (x2n−1)\tan \,\,x + \frac{1}{2}\,\,\tan \,\,\frac{x}{2}\,\, + \,\,\frac{1}{{{2^2}}}\,\,\tan \,\,\frac{x}{{{2^2}}} + \ldots + \frac{1}{{{2^{n - 1}}}}\,\,\tan \,\,\left( {\frac{x}{{{2^{n - 1}}}}} \right) is equal to
View answer and explanation
Correct option: B
12n−1  cot⁡  (x2n−1)−2cot⁡  2x\frac{1}{{{2^{n - 1}}}}\,\,\cot \,\,\left( {\frac{x}{{{2^{n - 1}}}}} \right) - 2\cot \,\,2x
Option B is correct because by using the identity tan⁡x=cot⁡x−2cot⁡2x\tan x = \cot x - 2\cot 2x repeatedly, we can express the given sum as a telescoping series, which simplifies to the given expression.

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46 Medium
46 of 122
Application +4 | -1
The value of sec⁡8θ−1sec⁡4θ−1\frac{\sec 8 \theta-1}{\sec 4 \theta-1} is
View answer and explanation
Correct option: C
tan⁡8θtan⁡2θ\frac{\tan 8 \theta}{\tan 2 \theta}
Option C (tan⁡8θtan⁡2θ\frac{{\tan 8 \theta}}{{\tan 2 \theta}}) is correct because the expression simplifies to tan⁡8θtan⁡2θ\frac{{\tan 8 \theta}}{{\tan 2 \theta}} using trigonometric identities.
47 Medium
47 of 122
Application +4 | -1
If k=sin⁡π/18sin⁡5π′18sin⁡7π/18k=\sin \pi / 18 \sin 5 \pi^{\prime} 18 \sin 7 \pi / 18, then the numerical value of kk is equal to
View answer and explanation
Correct option: C
1/81 / 8
Option C (1/81 / 8) is correct because it is obtained by using the product-to-sum trigonometric identities to simplify the expression.
48 Medium
48 of 122
Application +4 | -1
α,  β(α≠β)\alpha ,\;\beta \left( {\alpha \ne \beta } \right) satisfy the equation a  cos⁡θ+b  sin⁡θ=ca\,\,\cos \theta + b\,\,\sin \theta = c, then the value of tan⁡(α+β2)\tan \left( {\frac{{\alpha + \beta }}{2}} \right), is

View answer and explanation
Correct option: A
b/ab/a
Option A (b/ab/a) is correct because it represents the value of tan⁡(α+β2)\tan \left( {\frac{{\alpha + \beta }}{2}} \right) derived from transforming the given equation and using trigonometric identities.
49 Hard
49 of 122
Understanding +4 | -1
Let nn be an odd integer. If sin⁡  nθ=  ∑r=0n  br  sin⁡r  θ  \sin \,\,n\mathrm{\theta } = \mathrm{\;}\sum \limits_{r = 0}^n \,\,{b_r}\,\,{\sin ^r}\,\,\mathrm{\theta \;} for all real θ\mathrm{\theta }, then
View answer and explanation
Correct option: B
b0=0,   b1=n{b_{0 = }}0,\,\;{b_1} = n
Option B (b0=0,  b1=n{b_{0 = }}0,\;{b_1} = n) is correct because when θ=0\mathrm{\theta } = 0, sin⁡nθ=0\sin n\mathrm{\theta } = 0 and the derivative at θ=0\mathrm{\theta } = 0 gives b1=n{b_1} = n.
50 Easy
50 of 122
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 }.
51 Medium
51 of 122
Application +4 | -1
If a2−2acos⁡x+1=674a^2-2 a \cos x+1=674 and tan⁡(x/2)=7\tan (x / 2)=7 then the integral value of aa is
View answer and explanation
Correct option: A
25
Option A (25) is correct because solving the equation a2−2acos⁡x+1=674a^2-2 a \cos x+1=674 with tan⁡(x/2)=7\tan (x / 2)=7 yields a=25a=25.
52 Easy
52 of 122
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.
53 Easy
53 of 122
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.
54 Medium
54 of 122
Application +4 | -1
If α,  β\mathrm{\alpha },\mathrm{\;\beta } are different values of xx satisfying acos⁡x+bsin⁡x=ca\cos x + b\sin x = c, then tan⁡(  α+β2)\tan \left( {\;\frac{{\mathrm{\alpha } + \mathrm{\beta }}}{2}} \right) is equal to

View answer and explanation
Correct option: C
ba\frac{b}{a}
Option C (ba\frac{b}{a}) is correct because it is derived by using the tangent half-angle substitution and solving the resulting quadratic equation.
55 Easy
55 of 122
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.
56 Medium
56 of 122
Application +4 | -1
If cos⁡  20∘=k\cos \,\,20^\circ = k and cos   x=2k2−1cos\;\,x = 2{k^2} - 1, then the possible values of xx between 0∘0^\circ and 360∘360^\circ are
View answer and explanation
Correct option: C
40∘40^\circ and 320∘320^\circ
Option C (40∘40^\circ and 320∘320^\circ) is correct because using the double angle formula for cosine, we find that cos(x) = cos(40), leading to x = 40 and x = 360 - 40 = 320.
57 Easy
57 of 122
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.
58 Easy
58 of 122
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.
59 Medium
59 of 122
Application +4 | -1
If sin⁡θ1−sin⁡θ2=a\sin \theta_1-\sin \theta_2=a and cos⁡θ1+cos⁡θ2=b\cos \theta_1+\cos \theta_2=b, then
View answer and explanation
Correct option: B
a2+b2≤4\mathrm{a}^2+\mathrm{b}^2 \leq 4
Option B (a2+b2≤4a^2+b^2 \leq 4) is correct because by squaring and adding the given equations, we can simplify the expression to a2+b2=2+2cos⁡(θ1+θ2)a^2 + b^2 = 2 + 2 \cos(\theta_1 + \theta_2), which is always less than or equal to 4.
60 Easy
60 of 122
Application +4 | -1
The value of tan⁡  67  1∘2+cot⁡  67  1∘2\tan \,\,67\,\,\frac{{1^\circ }}{2} + \cot \,\,67\,\,\frac{{1^\circ }}{2} is
View answer and explanation
Correct option: C
2 22\,\sqrt 2
Option C (222\sqrt 2) is correct because simplifying the expression using trigonometric identities leads to the result 222\sqrt 2.
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