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JEE • MATHEMATICS • COMPOUND ANGLES

Compound angles practice · Mixed difficulty

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Showing 91-120 of 147 questions
91 Easy
91 of 147
Analyzing +4 | -1
If cos⁡  (θ1−θ2)cos⁡  (θ1+θ2)+cos⁡  (θ3+θ4)cos⁡  (θ3−θ4)=0\frac{{\cos \,\,\left( {{\theta _1} - {\theta _2}} \right)}}{{\cos \,\,\left( {{\theta _1} + {\theta _2}} \right)}} + \frac{{\cos \,\,\left( {{\theta _3} + {\theta _4}} \right)}}{{\cos \,\,\left( {{\theta _3} - {\theta _4}} \right)}} = 0, then tan⁡  θ1  tan⁡  θ2  tan⁡  θ3  tan⁡  θ4=\tan \,\,{\theta _1}\,\,\tan \,\,{\theta _2}\,\,\tan \,\,{\theta _3}\,\,\tan \,\,{\theta _4} =
View answer and explanation
Correct option: C
−1- 1
Option C (-1) is correct because the given condition leads to the relationship tan⁡θ1tan⁡θ2tan⁡θ3tan⁡θ4=−1\tan \theta_1 \tan \theta_2 \tan \theta_3 \tan \theta_4 = -1.
92 Easy
92 of 147
Application +4 | -1
If A+B=π4A + B = \frac{\pi }{4}, then (tan⁡A+1)  (tan⁡B+1)\left( {\tan A + 1} \right)\,\,\left( {\tan B + 1} \right) is equal to
View answer and explanation
Correct option: B
2
Option B (2) is correct because by using the tangent addition formula and the given condition, the expression simplifies to 2.
93 Easy
93 of 147
Application +4 | -1
If A,B,C,D\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D} are the angles of a quadrilaterals then the value of tan⁡A+tan⁡B+tan⁡C+tan⁡Dcot⁡A+cot⁡B+cot⁡C+cot⁡D\frac{\tan A+\tan B+\tan C+\tan D}{\cot A+\cot B+\cot C+\cot D} is
View answer and explanation
Correct option: A
tan⁡Atan⁡Btan⁡Ctan⁡D\tan A \tan B \tan C \tan D
Option A (tan⁡Atan⁡Btan⁡Ctan⁡D\tan A \tan B \tan C \tan D) is correct because it is derived from the trigonometric identities related to the sum of angles in a quadrilateral.
94 Medium
94 of 147
Application +4 | -1
tan⁡15∘+1tan⁡165∘+1tan⁡105∘+tan⁡195∘=2a\tan 15^{\circ}+\frac{1}{\tan 165^{\circ}}+\frac{1}{\tan 105^{\circ}}+\tan 195^{\circ}=2 a. Then the value of (a+1a)\left(a+\frac{1}{a}\right) is___________
View answer and explanation
Correct option: B
−43-\frac{4}{\sqrt{3}}
Option B (-43\frac{{4}}{{\sqrt{{3}}}}) is correct because it is the simplified value of a+1aa + \frac{{1}}{{a}} after calculating aa from the given trigonometric expression.
95 Easy
95 of 147
Analyzing +4 | -1
If angle θ\theta be divided into two parts such that the tangent of one part is kk times the tangent of the other and ϕ\phi is their difference, then sinθ\mathrm{sin \theta } is equal to
View answer and explanation
Correct option: A
k+1k−1  sin⁡  ϕ\frac{{k + 1}}{{k - 1}}\,\,\sin \,\,\phi
Option A (k+1k−1 sin⁡ ϕ\frac{{k + 1}}{{k - 1}}\,\sin \,\phi) is correct because by using componendo and dividendo rule on the given condition and applying trigonometric identities, we can express sin⁡θ\sin \theta in terms of kk and sin⁡ϕ\sin \phi.
96 Medium
96 of 147
Analyzing +4 | -1
If tan⁡  α+tan⁡  βcot⁡  α+cot⁡  β  +  {cos⁡  (α−β)  sec⁡  (α+β)+1}−1=1\frac{{\tan \,\,\alpha + \tan \,\,\beta }}{{\cot \,\,\alpha + \cot \,\,\beta }}\,\, + \,\,{\left\{ {\cos \,\,\left( {\alpha - \beta } \right)\,\,\sec \,\,\left( {\alpha + \beta } \right) + 1} \right\}^{ - 1}} = 1, then tan⁡α tan⁡  α  tan⁡  β\tan \,\,\alpha \,\,\tan \,\,\beta is equal to
View answer and explanation
Correct option: A
1
The correct answer is A (1) because simplifying the given trigonometric expression leads to tan⁡αtan⁡β=1\tan \alpha \tan \beta = 1.
97 Easy
97 of 147
Application +4 | -1
If α+β=π/2\alpha+\beta=\pi / 2 and β+γ=α\beta+\gamma=\alpha, then tan⁡α\tan \alpha equals
View answer and explanation
Correct option: C
tan⁡β+2tan⁡γ\tan \beta+2 \tan \gamma
Option C (tan⁡β+2tan⁡γ\tan \beta+2 \tan \gamma) is correct because by using the given relations and trigonometric identities, tan⁡α\tan \alpha can be expressed as tan⁡β+2tan⁡γ\tan \beta + 2 \tan \gamma.
98 Easy
98 of 147
Application +4 | -1
If tan⁡θ+tan⁡(π3+θ)+tan⁡(−π3+θ)=Ktan⁡3  θ\tan \theta + \tan \left( {\frac{\pi }{3} + \theta } \right) + \tan \left( {\frac{{ - \pi }}{3} + \theta } \right) = K\tan 3\;\theta, then the value of KK is
View answer and explanation
Correct option: C
3
Option C (3) is correct because simplifying the given trigonometric expression using tangent addition formulas leads to 3tan⁡3θ3\tan 3\theta, thus K = 3.
99 Easy
99 of 147
Understanding +4 | -1
The value of
cos⁡(45∘−A)cos⁡(45∘−B)−sin⁡(45∘−A)sin⁡(45∘−B)\cos \left(45^{\circ}-A\right) \cos \left(45^{\circ}-B\right)-\sin \left(45^{\circ}-A\right)\sin \left(45^{\circ}-B\right) is
View answer and explanation
Correct option: A
sin⁡(A+B)\sin (A+B)
Option B is correct because using the cosine addition formula, the expression simplifies to cos(90 - (A+B)) which equals sin(A+B).
100 Easy
100 of 147
Application +4 | -1
The value of   sin⁡  A  sin⁡(60∘−A)  sin⁡  (60∘+A)\mathrm{\;}\sin \,\,A\,\,\sin \left( {60^\circ - A} \right)\,\,\sin \,\,\left( {60^\circ + A} \right) is equal to
View answer and explanation
Correct option: C
sin⁡  3A4\frac{{\sin \,\,3A}}{4}
Option C (sin⁡ 3A4\frac{{\sin \,3A}}{4}) is correct because it is the result of applying the product-to-sum trigonometric identities.
101 Easy
101 of 147
Analyzing +4 | -1
In a △ABC\triangle A B C, if aa is the arithmetic mean and b,cb, c are two geometric means between any two positive number. Then, sin⁡3B+sin⁡3Csin⁡Asin⁡Bsin⁡C\frac{\sin ^3 B+\sin ^3 C}{\sin A \sin B \sin C} is equal to
View answer and explanation
Correct option: A
2
Option A (2) is correct because after substituting the given conditions and simplifying, the expression equals 2.
102 Easy
102 of 147
Application +4 | -1
The numerical value of tan⁡20∘tan⁡80∘cot⁡50∘\tan 20^{\circ} \tan 80^{\circ} \cot 50^{\circ} is
View answer and explanation
Correct option: A
3\sqrt{3}
Option A (3\sqrt{{3}}) is correct because the expression simplifies to 3\sqrt{{3}} using trigonometric identities.
103 Easy
103 of 147
Application +4 | -1
tan⁡  2π5−  tan⁡  π15−3tan⁡  2π5  tan⁡  π15\tan \,\,\frac{{2\pi }}{5} - \,\,\tan \,\,\frac{\pi }{{15}} - \sqrt 3 \tan \,\,\frac{{2\pi }}{5}\,\,\tan \,\,\frac{\pi }{{15}} is equal to
View answer and explanation
Correct option: D
3\sqrt 3
Option D (3\sqrt 3) is correct because the expression simplifies to tan⁡(2π5−π15)=tan⁡(π3)=3\tan(\frac{{2\pi }}{5} - \frac{\pi }{{15}}) = \tan(\frac{\pi}{3}) = \sqrt{3} using the tangent subtraction formula.
104 Easy
104 of 147
Application +4 | -1
If sin⁡  x+cos⁡  x=15\sin \,\,x + \cos \,\,x = \frac{1}{5}, then tan⁡  2x\tan \,\,2x is
View answer and explanation
Correct option: B
247\frac{{24}}{{7}}
Option B (247\frac{{24}}{{7}}) is correct because by squaring the given equation and using trigonometric identities, we can find sin⁡2x\sin 2x and cos⁡2x\cos 2x, and then calculate tan⁡2x\tan 2x.
105 Medium
105 of 147
Analyzing +4 | -1
If x+y−z+xyz=0x + y - z + xyz = 0, then 2x1−x2+2y1−y2−2z1−z2\frac{{2x}}{{1 - {x^2}}} + \frac{{2y}}{{1 - {y^2}}} - \frac{{2z}}{{1 - {z^2}}} is equal to
View answer and explanation
Correct option: C
8xyz[(1−x2)(1−y2)(1−z2)]\frac{8 x y z}{\left[\left(1-x^2\right)\left(1-y^2\right)\left(1-z^2\right)\right]}
Option D (−8xyz[(1−x2)(1−y2)(1−z2)]\frac{{-8 x y z}}{{\left[(1-x^2)(1-y^2)(1-z^2)\right]}}) is the correct answer because it is derived using trigonometric substitutions and the tangent addition formula.

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106 Medium
106 of 147
Application +4 | -1
If tan⁡x+tan⁡(x+π/3)+\tan x+\tan (x+\pi / 3)+ tan⁡(x+2π/3)=3\tan (x+2 \pi / 3)=3, then
View answer and explanation
Correct option: C
tan⁡3x=1\tan 3 x=1
Option C (tan⁡3x=1\tan 3x = 1) is correct because simplifying the given trigonometric equation leads to the result tan⁡3x=1\tan 3x = 1.
107 Medium
107 of 147
Application +4 | -1
The value of ∑r=15cos⁡(2r−1)π11\sum\limits_{{r=1}}^5 \cos (2 r-1) \frac{\pi}{11} is
View answer and explanation
Correct option: A
12\frac{1}{2}
Option A (12\frac{1}{2}) is correct because using the sum-to-product formula and simplifying the expression leads to this result.
108 Easy
108 of 147
Application +4 | -1
If 0<α,β,γ<π/20<\alpha, \beta, \gamma<\pi / 2 such that α+β+γ=π2\alpha+\beta+\gamma=\frac{\pi}{2} and cot⁡α,cot⁡β,cot⁡\cot \alpha, \cot \beta, \cot γ\gamma are in arithmetic progression, then the value of cot⁡αcot⁡γ\cot \alpha \cot \gamma is
View answer and explanation
Correct option: B
3
The correct answer is A (n+m+1Cn+1{ }^{n+m+1} \mathrm{C}_{n+1}) because it is the result of applying the Hockey Stick Identity to the given summation.
109 Easy
109 of 147
Application +4 | -1
If cos⁡  α+cos⁡  β=0=sin⁡  α+sin⁡  β\cos \,\,\alpha + \cos \,\,\beta = 0 = \sin \,\,\alpha + \sin \,\,\beta, then cos⁡  2  α+cos⁡  2  β=\cos \,\,2\;\alpha + \cos \,\,2\;\beta =
View answer and explanation
Correct option: B
− 2  cos⁡  (α+β)- \,2\,\,\cos \,\,\left( {\alpha + \beta } \right)
Option B (−2cos⁡(α+β)-2 \cos(\alpha + \beta)) is correct because it is derived by squaring the given equations and using trigonometric identities.
110 Easy
110 of 147
Application +4 | -1
If α\alpha and β\beta satisfying 2  sec⁡  2α=tan⁡  β+cot⁡  β2\,\,\sec \,\,2\alpha = \tan \,\,\beta + \cot \,\,\beta, then α+β\alpha + \beta is equal to
View answer and explanation
Correct option: C
π4\frac{\pi }{4}
Option C (π4\frac{\pi}{4}) is correct because it is the value of α+β\alpha + \beta that satisfies the given trigonometric equation.
111 Easy
111 of 147
Application +4 | -1
If tan⁡15∘+1tan⁡75∘+1tan⁡105∘+tan⁡195∘=2a\tan 15^{\circ}+\frac{1}{\tan 75^{\circ}}+\frac{1}{\tan 105^{\circ}}+\tan 195^{\circ}=2 a, then the value of (a+1a)\left(a+\frac{1}{a}\right) is
View answer and explanation
Correct option: A
4
Option A (4) is correct because simplifying the trigonometric expression results in 2*tan(15), which leads to a = 2-sqrt(3), and a + 1/a = 4.
112 Easy
112 of 147
Application +4 | -1
cos⁡  α  sin⁡  (β−γ)+cos⁡  β  sin⁡  (γ−α)+cos⁡  γ  sin⁡  (α−β)\cos \,\,\alpha \,\,\sin \,\,\left( {\beta - \gamma } \right) + \cos \,\,\beta \,\,\sin \,\,\left( {\gamma - \alpha } \right) + \cos \,\,\gamma \,\,\sin \,\,\left( {\alpha - \beta } \right) is equal to
View answer and explanation
Correct option: A
0\mathrm{0}
Option A (0) is correct because the given trigonometric expression simplifies to zero using trigonometric identities.
113 Medium
113 of 147
Application +4 | -1
The value of 16cos⁡(2π15)⋅cos⁡(4π15)cos⁡(8π15)⋅cos⁡(16π15)16 \cos \left(\frac{2 \pi}{15}\right) \cdot \cos \left(\frac{4 \pi}{15}\right) \cos \left(\frac{8 \pi}{15}\right) \cdot \cos \left(\frac{16 \pi}{15}\right) is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the expression simplifies to 1 using trigonometric identities.
114 Medium
114 of 147
Analyzing +4 | -1
If α,   β,   γ∈  [0,   π2]\alpha ,\;\,\beta ,\;\,\gamma \in \,\,\left[ {0,\,\;\frac{\pi }{2}} \right], then the value of sin⁡  (α+β+γ)sin⁡  α  +  sin⁡  β  +  sin⁡  γ\frac{{\sin \,\,\left( {\alpha + \beta + \gamma } \right)}}{{\sin \,\,\alpha \,\, + \,\,\sin \,\,\beta \,\, + \,\,\sin \,\,\gamma }} is

View answer and explanation
Correct option: A
<1< 1
Option A (<1< 1) is correct because the sine of the sum of angles is always less than the sum of the sines of the individual angles in the given interval.
115 Easy
115 of 147
Application +4 | -1
If     π2<α<π,  π<β<  3π2;  sin⁡  α=1517    and  tan⁡  β=125\;\;\frac{\pi }{2} < \alpha < \pi ,\,\,\pi < \beta < \,\,\frac{{3\pi }}{2};\,\,\sin \,\,\alpha = \frac{{15}}{{17}}\,\,\;\mathrm{and}\,\,\tan \,\,\beta = \frac{{12}}{5} then the value of sin⁡ (β−α)\sin \,\left( {\beta - \alpha } \right) is
View answer and explanation
Correct option: D
171221\frac{{171}}{{221}}
Option D (171221\frac{{171}}{{221}}) is correct because, by using the given trigonometric values and the angle subtraction formula, we can calculate the value of sin⁡(β−α)\sin(\beta - \alpha).
116 Easy
116 of 147
Application +4 | -1
If   tan⁡A−tan⁡  B=x   and  cot⁡  B−cot⁡  A=y,  then  cot⁡(A−B)\;\tan A - \tan \,\,B = x\,\mathrm{\;and}\,\,\cot \,\,B - \cot \,\,A = y,\mathrm{\;then}\,\,\cot \left( {A - B} \right) is equal to

View answer and explanation
Correct option: D
1x+1y\frac{1}{x} + \frac{1}{y}
Option D (1x+1y\frac{1}{x} + \frac{1}{y}) is correct because it represents the correct expression for cot⁡(A−B)\cot(A - B) in terms of xx and yy.
117 Easy
117 of 147
Application +4 | -1
If sin⁡  A=110    \sin \,\,A = \frac{1}{{\sqrt {10\;} }}\; and sin⁡  B=15  \sin \,\,B = \frac{1}{{\sqrt 5 \;}}, where A  A\; and B  B\; are positive acute angles, then (A+B )(A + B\,) is equal to
View answer and explanation
Correct option: D
π4\frac{\pi }{4}
Option D (π4\frac{\pi }{4}) is correct because the sum of angles A and B is calculated to be π4\frac{\pi }{4} using trigonometric identities.
118 Medium
118 of 147
Application +4 | -1
If sin⁡2x=α−1\sin 2 x=\alpha-1 and cos⁡2x=β−1\cos 2 x=\beta-1, then the value of sec⁡2x[(cos⁡2x−sin⁡2x)−2sin⁡xcos⁡x]1+sin⁡2x\frac{\sec ^2 x\left[\left(\cos ^2 x-\sin ^2 x\right)-2 \sin x \cos x\right]}{1+\sin 2 x} is equal to
View answer and explanation
Correct option: B
2α−2β\frac{2}{\alpha}-\frac{2}{\beta}
Option C (2β−2α\frac{2}{\beta}-\frac{2}{\alpha}) is correct because the expression simplifies to 2(1α−1β)2(\frac{1}{\alpha} - \frac{1}{\beta}).
119 Easy
119 of 147
Application +4 | -1
If cos⁡  (α+β)=45,  sin⁡  (α−β)=513\cos \,\,\left( {\alpha + \beta } \right) = \frac{4}{5},\,\,\sin \,\,\left( {\alpha - \beta } \right) = \frac{5}{{13}} and α,   β\alpha ,\;\,\beta lie between 0\mathrm{0} and π4\frac{\pi }{4}, then tan⁡  2α=\tan \,\,2\alpha =

View answer and explanation
Correct option: B
3356\frac{{33}}{{56}}
Option A is correct because by using the given information to find tan⁡(α+β)\tan(\alpha + \beta) and tan⁡(α−β)\tan(\alpha - \beta), we can then find tan⁡2α\tan 2\alpha using the tangent addition formula.
120 Easy
120 of 147
Analyzing +4 | -1
If sin⁡A+sin⁡  B=a\sin A + \sin \;B = a and cos⁡A+cos⁡B=b\cos A + \cos B = b, then cos⁡(A+B)\cos \left( {A + B} \right)
View answer and explanation
Correct option: A
a2+b2b2−a2\frac{{{a^2} + {b^2}}}{{{b^2} - {a^2}}}
Option C (b2−a2a2+b2\frac{{{b^2} - {a^2}}}{{{a^2} + {b^2}}}) is correct because by squaring and manipulating the given equations, we can express cos⁡(A+B)\cos(A+B) in terms of aa and bb.
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