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JEE • MATHEMATICS • COMPOUND ANGLES

Compound angles practice · Mixed difficulty

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Showing 121-147 of 147 questions
121 Easy
121 of 147
Analyzing +4 | -1
If tan⁡θ+tan⁡(π3+θ)+tan⁡(−π3+θ)=atan⁡3θ,\tan \mathrm{\theta } + \tan \left( {\frac{\pi }{3} + \mathrm{\theta }} \right) + \tan \left( { - \frac{\pi }{3} + \mathrm{\theta }} \right) = a\tan 3\mathrm{\theta }, then aa is equal to
View answer and explanation
Correct option: C
3
Option C (3) is correct because simplifying the trigonometric expression leads to 3tan⁡3θ=atan⁡3θ3\tan 3\theta = a\tan 3\theta, thus a=3a = 3.
122 Easy
122 of 147
Analyzing +4 | -1
If m  tan⁡  (θ−30∘)=n  tan⁡  (θ+120∘)m\,\,\tan \,\,\left( {\theta - 30^\circ } \right) = n\,\,\tan \,\,\left( {\theta + 120^\circ } \right), then cos⁡  2θ\cos \,\,2\theta equals
View answer and explanation
Correct option: D
m+n2(m−n)\frac{{m + n}}{{2\left( {m - n} \right)}}
Option D (m+n2(m−n)\frac{{m + n}}{{2\left( {m - n} \right)}}) is correct because it is the result of simplifying the trigonometric equation and solving for cos⁡2θ\cos 2\theta.
123 Easy
123 of 147
Application +4 | -1
If tan⁡  A\tan \,\,A and tan⁡  B\tan \,\,B are the roots of abx2−c2x+ab=0ab{x^2} - {c^2}x + ab = 0 where a,   b,   ca,\,\;b,\,\;c are the sides of the triangles ABCABC, then the value of sin⁡2A+sin⁡2B+sin⁡2C{\sin ^2}A + {\sin ^2}B + {\sin ^2}C is
View answer and explanation
Correct option: D
2
Option D (2) is correct because, given the roots of the quadratic equation, we can deduce that the triangle is right-angled, and then calculate the value of the expression.
124 Medium
124 of 147
Analyzing +4 | -1
If sin⁡(θ+α)=a\sin (\theta+\alpha)=a and sin⁡(θ+β)=b\sin (\theta+\beta)=b, then cos⁡2(α−β)−4abcos⁡(α−β)\cos 2(\alpha-\beta)-4 a b \cos (\alpha-\beta) is equal to:
View answer and explanation
Correct option: B
1−2a2−2b21-2 a^2-2 b^2
Option B (1−2a2−2b21-2 a^2-2 b^2) is correct because by using trigonometric identities and the given equations, the expression simplifies to this result.
125 Easy
125 of 147
Analyzing +4 | -1
In a Δ PQR,   ∠R=π2\Delta \,PQR,\,\;\angle R = \frac{\pi }{2}. If tan⁡  P2\tan \,\,\frac{P}{2} and tan⁡  Q2\tan \,\,\frac{Q}{2} are the roots of the equation ax2+  bx+c=0(a≠0)a{x^2} + \,\,bx + c = 0\left( {a \ne 0} \right), then
View answer and explanation
Correct option: A
a+b=ca + b = c
Option A (a+b=ca + b = c) is correct because it is derived from the relationship between the roots of the quadratic equation and the fact that P+Q=π2P + Q = \frac{\pi}{2} in the given triangle.
126 Hard
126 of 147
Analyzing +4 | -1
If
α,   β,   γ∈(0,   π/2)\alpha ,\,\;\beta ,\,\;\gamma \in \left( {0,\;\,\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 sin⁡α+sin⁡β+sin⁡γ>sin⁡(α+β+γ)\sin \alpha + \sin \beta + \sin \gamma > \sin (\alpha + \beta + \gamma) when α,β,γ∈(0,π/2)\alpha, \beta, \gamma \in (0, \pi/2).
127 Medium
127 of 147
Application +4 | -1
The value of the expression 1cos⁡290∘+13sin⁡250∘\frac{1}{\cos 290^{\circ}}+\frac{1}{\sqrt{3} \sin 250^{\circ}} is equal to
View answer and explanation
Correct option: B
4/34 / \sqrt{3}
Option B (4/34 / \sqrt{3}) is correct because the expression simplifies to 4/34/\sqrt{3} using trigonometric identities and angle transformations.
128 Easy
128 of 147
Application +4 | -1
If cos⁡  (θ−α)=a,  cos⁡  (θ−β)=b\cos \,\,\left( {\theta - \alpha } \right) = a,\,\,\cos \,\,\left( {\theta - \beta } \right) = b, then sin⁡2  (α−β)+2 ab  cos⁡  (α−β){\sin ^2}\,\,\left( {\alpha - \beta } \right) + 2\,ab\,\,\cos \,\,\left( {\alpha - \beta } \right) is equal to
View answer and explanation
Correct option: A
a2+b2{a^2} + {b^2}
Option A (a2+b2{a^2} + {b^2}) is correct because it is the simplified result after applying trigonometric identities and algebraic manipulation to the given equations.
129 Hard
129 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 sin⁡(α+β+γ)<sin⁡α+sin⁡β+sin⁡γ\sin(\alpha + \beta + \gamma) < \sin \alpha + \sin \beta + \sin \gamma for α,β,γ∈(0,π/2)\alpha, \beta, \gamma \in (0, \pi/2).
130 Hard
130 of 147
Analyzing +4 | -1
If pp is the product of the sines of angles of a triangle, and qq the product of their cosines, then tangents of the angles are roots of the equation
View answer and explanation
Correct option: A
qx3−px3+(1+q)x−p=0q{x^3} - p{x^3} + \left( {1 + q} \right)x - p = 0
Option A (qx3−px3+(1+q)x−p=0q{x^3} - p{x^3} + \left( {1 + q} \right)x - p = 0) is correct because the tangents of the angles are the roots of this equation, derived from Vieta's formulas and trigonometric identities.
131 Medium
131 of 147
Application +4 | -1
If tan⁡α,tan⁡β,tan⁡γ\tan \alpha, \tan \beta, \tan \gamma are the roots of the equation x3−px2−r=0x^3-p x^2-r=0, then the value of (1+tan⁡2α)(1+tan⁡2β)(1+tan⁡2γ)\left(1+\tan ^2 \alpha\right)\left(1+\tan ^2 \beta\right)\left(1+\tan ^2 \gamma\right) is equal to
View answer and explanation
Correct option: B
1+(p−r)21+(p-r)^2
Option B (1+(p−r)21+(p-r)^2) is correct because it is the result of expanding and simplifying the expression (1+tan⁡2α)(1+tan⁡2β)(1+tan⁡2γ)(1+\tan ^2 \alpha)(1+\tan ^2 \beta)(1+\tan ^2 \gamma) using Vieta's formulas.
132 Easy
132 of 147
Analyzing +4 | -1
In △ABC\triangle A B C, line joining the circumcentre and orthocentre is parallel to side ACA C, then the value of tan⁡A⋅tan⁡C\tan A \cdot \tan C is equal to
View answer and explanation
Correct option: B
3
Option B (3) is correct because by equating the distance of the circumcentre and orthocentre from side AC, we can derive the relation tan⁡A⋅tan⁡C=3\tan A \cdot \tan C = 3.
133 Easy
133 of 147
Application +4 | -1
If tan⁡α=1/7\tan \alpha=1 / 7 and sin⁡β=1/10\sin \beta=1 / \sqrt{10} where 0<α,β<π/20<\alpha, \beta<\pi / 2, then 2β2 \beta is equal to
View answer and explanation
Correct option: A
π/4−α\pi / 4-\alpha
Option A (π4−α\frac{\pi}{4} - \alpha) is correct because by finding tan⁡2β\tan 2\beta and then tan⁡(α+2β)\tan (\alpha + 2\beta), we can determine that α+2β=π4\alpha + 2\beta = \frac{\pi}{4}, leading to 2β=π4−α2\beta = \frac{\pi}{4} - \alpha.
134 Medium
134 of 147
Application +4 | -1
If sin⁡β\sin \beta is the GM between sin⁡α\sin \alpha and cos⁡α\cos \alpha, then cos⁡  2β=\cos \,\,2\beta =
View answer and explanation
Correct option: C
cos⁡2  (π4+α){\cos ^2}\,\,\left( {\frac{\pi }{4} + \alpha } \right)
Option B (2cos⁡2  (π4+α)2{\cos ^2}\,\,\left( {\frac{\pi }{4} + \alpha } \right)) is incorrect. Option C (cos⁡2  (π4+α){\cos ^2}\,\,\left( {\frac{\pi }{4} + \alpha } \right)) is correct because simplifying the expression cos⁡ 2β\cos \,2\beta using the given GM condition yields 2cos⁡2  (π4+α)2{\cos ^2}\,\,\left( {\frac{\pi }{4} + \alpha } \right).
135 Easy
135 of 147
Application +4 | -1
If   tan α,  tan⁡ β,  tan⁡ γ  \mathrm{\;tan}\,\alpha ,\,\,\tan \,\beta ,\,\,\tan \,\gamma \; are the roots of the equation x3−px2−r=0{x^3} - p{x^2} - r = 0, then the value of (1+tan⁡2α)  (1+tan⁡2β)  (1+tan⁡2γ)\left( {1 + {{\tan }^2}\alpha } \right)\,\,\left( {1 + {{\tan }^2}\beta } \right)\,\,\left( {1 + {{\tan }^2}\gamma } \right) is equal to
View answer and explanation
Correct option: B
1+(p−r)21 + {\left( {p - r} \right)^2}
Option B is correct because it represents the simplified expression after applying Vieta's formulas and trigonometric identities to the given equation.

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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 Medium
136 of 147
Application +4 | -1
If tan⁡θ1,tan⁡θ2,tan⁡θ3\tan \theta_1, \tan \theta_2, \tan \theta_3 and tan⁡θ4\tan \theta_4 are the roots of the equation

x4−x3sin⁡2β+x2cos⁡2β−xcos⁡β−sin⁡β=0x^4-x^3 \sin 2 \beta+x^2 \cos 2 \beta-x \cos \beta-\sin \beta=0
then tan⁡(θ1+θ2+θ3+θ4)\tan \left(\theta_1+\theta_2+\theta_3+\theta_4\right) is equal to
View answer and explanation
Correct option: D
cot⁡β\cot \beta
Option D (cot⁡β\cot \beta) is correct because it is obtained by using Vieta's formulas to find the sum and product of roots, and then applying the tangent addition formula.
137 Easy
137 of 147
Analyzing +4 | -1
If tan⁡  θ  tan⁡(π3+θ)  tan⁡  (−π3+θ)=k  tan⁡  3θ\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: D
None of these
Option D (None of these) is correct because the simplification of the given expression results in k=−1k = -1, which is not among the provided options.
138 Medium
138 of 147
Application +4 | -1
If     sin⁡  xsin⁡  y=12, cos⁡  xcos⁡  y=32\;\;\frac{{\sin \,\,x}}{{\sin \,\,y}} = \frac{1}{2},\,\frac{{\cos \,\,x}}{{\cos \,\,y}} = \frac{3}{2}, where x,   y∈(0,  π2)x,\;\,y \in \left( {0,\,\,\frac{\pi }{2}} \right), then the value of tan⁡  x+y\tan \,\,x + y is equal to

View answer and explanation
Correct option: D
15\sqrt {15}
Option D (15\sqrt {15}) is correct because it is the result of correctly applying trigonometric identities and solving for tan⁡ (x+y)\tan \,(x + y).
139 Easy
139 of 147
Application +4 | -1
If A+C=BA+C=B, then tan⁡Atan⁡Btan⁡C\tan A \tan B \tan C is equal to
View answer and explanation
Correct option: B
tan⁡B−tan⁡C−tan⁡A\tan B-\tan C-\tan A
Option B (tan B - tan C - tan A) is correct because it is derived from applying the tangent addition formula and rearranging the terms.
140 Medium
140 of 147
Application +4 | -1
Let A,  BA,\;B and CC are the angles of a teiangle and tan⁡(A2)=13,tan⁡(B2)=23\tan \left( {\frac{A}{2}} \right) = \frac{1}{3},\tan \left( {\frac{B}{2}} \right) = \frac{2}{3}. Then, tan⁡(C2)\tan \left( {\frac{C}{2}} \right) is equal to

View answer and explanation
Correct option: D
1 / 3\mathrm{1}\,\mathrm{/}\,\mathrm{3}
Option A (1 / 3\mathrm{1}\,\mathrm{/}\,\mathrm{3}) is incorrect. The correct answer is 7/9. The value of tan⁡(C2)\tan \left( {\frac{C}{2}} \right) is found using the identity tan⁡A2tan⁡B2+tan⁡B2tan⁡C2+tan⁡C2tan⁡A2=1\tan \frac{A}{2} \tan \frac{B}{2} + \tan \frac{B}{2} \tan \frac{C}{2} + \tan \frac{C}{2} \tan \frac{A}{2} = 1 in a triangle.
141 Easy
141 of 147
Application +4 | -1
sin⁡A+sin⁡  B=3  (cos⁡  B−cos⁡  A)⇒sin⁡  3A+sin⁡  3B\sin A + \sin \,\,B = \sqrt 3 \,\,\left( {\cos \,\,B - \cos \,\,A} \right) \Rightarrow \sin \,\,3A + \sin \,\,3B is equal to
View answer and explanation
Correct option: A
0\mathrm{0}
Option A (0\mathrm{0}) is correct because simplifying the given condition leads to A=−BA = -B, and thus sin⁡3A+sin⁡3B=0\sin 3A + \sin 3B = 0.
142 Easy
142 of 147
Application +4 | -1
If cos⁡(A−B)=35\cos \left( {A - B} \right) = \frac{3}{5} and tan⁡A  tan⁡B=2\tan A\,\,\tan B = 2, then
View answer and explanation
Correct option: A
cos⁡A  cos⁡B=15\cos A\,\,\cos B = \frac{1}{5}
Option A (cos A cos B = 1/5) is correct because, given the conditions, solving for tan A and tan B allows us to find cos A cos B = 1/5.
143 Easy
143 of 147
Application +4 | -1
If cot⁡α=1\cot \alpha=1 and sec⁡β=−53\sec \beta=-\frac{5}{3}, where π<α<3π2\pi<\alpha<\frac{3 \pi}{2} and π2<β<π\frac{\pi}{2}<\beta<\pi, then the value of tan⁡(α+\tan (\alpha+ β)\beta) and the quadrant in which α+β\alpha+\beta lies, respectively are :
View answer and explanation
Correct option: A
−17-\frac{1}{7} and IVth quadrant
The correct answer is A (-1/7 and IVth quadrant) because it correctly calculates the tangent of the sum of angles and identifies the correct quadrant.
144 Medium
144 of 147
Application +4 | -1
If A=35∘,  B=15∘A = 35^\circ ,\;B = 15^\circ and C=40∘C = 40^\circ, then tan⁡A .  tan⁡B+tan⁡B .  tan⁡C+tan⁡C . tan⁡A\tan A\,.\,\,\tan B + \tan B\,.\,\,\tan C + \tan C\,.\,\tan A is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because using the tangent addition formula for three angles, tan⁡(A+B+C)\tan(A + B + C), and knowing that A+B+C=90∘A + B + C = 90^\circ, we can derive that tan⁡A⋅tan⁡B+tan⁡B⋅tan⁡C+tan⁡C⋅tan⁡A=1\tan A \cdot \tan B + \tan B \cdot \tan C + \tan C \cdot \tan A = 1.
145 Easy
145 of 147
Application +4 | -1
The value of tan⁡  70∘−tan⁡  20∘tan⁡  50∘\frac{{\tan \,\,70^\circ - \tan \,\,20^\circ }}{{\tan \,\,50^\circ }} is equal to
View answer and explanation
Correct option: A
2
Option A (2) is correct because simplifying the trigonometric expression using the tangent addition formula and complementary angle identities results in the value 2.
146 Easy
146 of 147
Application +4 | -1
If A+B+C=0A + B + C = 0, then the value of ∑cot⁡(B+C−A)cot⁡(C+A−B)\sum \cot \left( {B + C - A} \right)\cot \left( {C + A - B} \right) is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because by using the condition A + B + C = 0, the expression simplifies to a known trigonometric identity which equals 1.
147 Easy
147 of 147
Analyzing +4 | -1
If tan⁡  θ+tan⁡  (θ+π3)+tan⁡  (θ+2π3)=3\tan \,\,\theta + \tan \,\,\left( {\theta + \frac{\pi }{3}} \right) + \tan \,\,\left( {\theta + \frac{{2\pi }}{3}} \right) = 3, then which of the following is equal to 1\mathrm{1} ?
View answer and explanation
Correct option: B
tan⁡  3θ\tan \,\,3\theta
The correct answer is B (tan⁡3θ\tan 3\theta) because simplifying the given equation using trigonometric identities leads to tan⁡3θ=1\tan 3\theta = 1.
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