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JEE • MATHEMATICS • COMPOUND ANGLES

Compound angles practice · Mixed difficulty

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Showing 31-60 of 147 questions
31 Medium
31 of 147
Application +4 | -1
If x=sin⁡α,y=sin⁡β,z=sin⁡(α+β)x=\sin \alpha, y=\sin \beta, z=\sin (\alpha+\beta) then cos⁡(α+β)=\cos (\alpha+\beta)=
View answer and explanation
Correct option: D
z2−x2−y2xy\frac{z^2-x^2-y^2}{x y}
Option C, z2−x2−y22xy\frac{z^2-x^2-y^2}{2 x y}, is the correct answer. It is derived by using the sine addition formula and trigonometric identities to express cos⁡(α+β)\cos(\alpha+\beta) in terms of xx, yy, and zz.
32 Easy
32 of 147
Application +4 | -1
If   α+β+γ=2π\;\alpha + \beta + \gamma = 2\pi, then
View answer and explanation
Correct option: A
tan⁡  α2+tan⁡  β2+tan⁡  γ2=tan⁡  α2  tan⁡  β2  tan⁡  γ2\tan \,\,\frac{\mathrm{\alpha }}{2} + \tan \,\,\frac{\mathrm{\beta }}{2} + \tan \,\,\frac{\mathrm{\gamma }}{2} = \tan \,\,\frac{\mathrm{\alpha }}{2}\,\,\tan \,\,\frac{\mathrm{\beta }}{2}\,\,\tan \,\,\frac{\mathrm{\gamma }}{2}
Option A is correct because when α+β+γ=2π\alpha + \beta + \gamma = 2\pi, the sum of the tangents of half angles equals their product.
33 Medium
33 of 147
Application +4 | -1
Given both θ\theta and ∅\emptyset are the acute angles sin⁡  θ=12,  cos⁡  ∅=13  \sin \,\,\theta = \frac{1}{2},\,\,\cos \,\,\emptyset = \frac{1}{3}\;, then the value of θ+∅\theta + \emptyset belongs to
View answer and explanation
Correct option: B
(π2,   2π3)\left( {\frac{\pi }{2},\,\;\frac{{2\pi }}{3}} \right)
Option B ((π2,   2π3\frac{\pi }{2},\,\;\frac{{2\pi }}{3})) is correct because, given the values of sin⁡θ\sin \theta and cos⁡ϕ\cos \phi, the sum θ+ϕ\theta + \phi falls within this interval.
34 Medium
34 of 147
Application +4 | -1
If sin⁡α+sin⁡β+sin⁡γ=0\sin \alpha+\sin \beta+\sin \gamma=0 and cos⁡α+\cos \alpha+ cos⁡β+cos⁡γ=0\cos \beta+\cos \gamma=0, then value of cos⁡(α−β)+cos⁡(β−γ)+\cos (\alpha-\beta)+\cos (\beta-\gamma)+ cos⁡(γ−α)\cos (\gamma-\alpha) is
View answer and explanation
Correct option: A
−3/2-3 / 2
Option A (-3/2) is correct because the given conditions lead to cos⁡(α−β)=cos⁡(β−γ)=cos⁡(γ−α)=−1/2\cos(\alpha-\beta) = \cos(\beta-\gamma) = \cos(\gamma-\alpha) = -1/2, and their sum is -3/2.
35 Easy
35 of 147
Application +4 | -1
If tan⁡A−tan⁡B=x\tan A-\tan B=x and cot⁡B−cot⁡A=y\cot B-\cot A=y, then cot⁡(A−B)\cot (A-B) is equal to
View answer and explanation
Correct option: C
1x+1y\frac{1}{x}+\frac{1}{y}
Option C (1x+1y\frac{{1}}{{x}} + \frac{{1}}{{y}}) is correct because cot⁡(A−B)\cot(A-B) can be expressed in terms of tan⁡A−tan⁡B\tan A - \tan B and cot⁡B−cot⁡A\cot B - \cot A, and then simplified.
36 Medium
36 of 147
Analyzing +4 | -1
If   tan⁡  α=1x(x2+x+1),  tan⁡  β=xx2+x+1  \mathrm{\;}\tan \,\,\mathrm{\alpha } = \frac{1}{{\sqrt {x\left( {{x^2} + x + 1} \right)} }},\,\,\tan \,\,\mathrm{\beta } = \frac{{\sqrt x }}{{\sqrt {{x^2} + x + 1} }}\mathrm{\;} and   tan⁡  γ=x−3+x−2+x−1,\mathrm{\;}\tan \,\,\mathrm{\gamma } = \sqrt {{x^{ - 3}} + {x^{ - 2}} + {x^{ - 1}},} then   α+β  \mathrm{\;\alpha } + \mathrm{\beta \;} is
View answer and explanation
Correct option: A
γ\gamma
Option A (γ\gamma) is correct because by using the tangent addition formula and simplifying, tan⁡(α+β)\tan(\alpha + \beta) equals tan⁡(γ)\tan(\gamma), thus α+β=γ\alpha + \beta = \gamma.
37 Easy
37 of 147
Application +4 | -1
If cos⁡θ+sin⁡θ=a,cos⁡2θ=b\cos \theta+\sin \theta=a, \cos 2 \theta=b, then
View answer and explanation
Correct option: B
b2=a2(2−a2)b^2=a^2\left(2-a^2\right)
Option B (b2=a2(2−a2)b^2=a^2(2-a^2)) is correct because squaring the first equation and using trigonometric identities leads to the given relationship between aa and bb.
38 Medium
38 of 147
Application +4 | -1
If ∣ k ∣=5\left| {\,k\,} \right| = 5 and 0∘  ≤θ≤360∘0^\circ \,\, \le \theta \le 360^\circ, then the number of different solutions of 3  cos⁡  θ+4  sin⁡  θ=k3\,\,\cos \,\,\theta + 4\,\,\sin \,\,\theta = k is
View answer and explanation
Correct option: B
Two
Option B (Two) is correct because the equation 3cosθ + 4sinθ = k has two solutions when |k| = 5, corresponding to the maximum and minimum values of the expression.
39 Easy
39 of 147
Application +4 | -1
sin⁡  120∘  cos⁡  150∘−  cos⁡  240∘  sin⁡  330∘\sin \,\,120^\circ \,\,\cos \,\,150^\circ - \,\,\cos \,\,240^\circ \,\,\sin \,\,330^\circ is equal to
View answer and explanation
Correct option: B
−1- 1
Option B (−1-1) is correct because the expression simplifies to −1-1 using trigonometric identities.
40 Easy
40 of 147
Application +4 | -1
If π<2θ<3π2\pi < 2 \theta < \frac{3 \pi}{2}, then 2+2+2cos⁡4θ\sqrt{2+\sqrt{2+2 \cos 4 \theta}} equals
View answer and explanation
Correct option: D
2sin⁡θ2 \sin \theta
Option D (2sin⁡θ2 \sin \theta) is correct because simplifying the expression using trigonometric identities and the given range for θ\theta leads to this result.
41 Medium
41 of 147
Application +4 | -1
If α+β=π2,   β+γ=α\alpha + \beta = \frac{\pi }{2},\;\,\beta + \gamma = \alpha, then the value of tan⁡α\tan \alpha equals

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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.
42 Easy
42 of 147
Application +4 | -1
If tan⁡(u+iv)=i\tan (u+i v)=i, then the value of vv is:
View answer and explanation
Correct option: B
∞\infty
Option B (∞\infty) is correct because for tan(u+iv) to be equal to i, v must approach infinity.
43 Easy
43 of 147
Application +4 | -1
If cos⁡(A−B)=3/5\cos \left( {A - B} \right) = 3/5 and tan⁡A  tan⁡B=2\tan A\,\,\tan B = 2, then which one of the following is true?
View answer and explanation
Correct option: D
cos⁡(A+B)=−15\cos \left( {A + B} \right) = - \frac{1}{5}
Option D (cos⁡(A+B)=−15\cos \left( {A + B} \right) = - \frac{1}{5}) is correct because using the given conditions cos⁡(A−B)=3/5\cos \left( {A - B} \right) = 3/5 and tan⁡A tan⁡B=2\tan A\,\tan B = 2, we can derive that cos⁡(A+B)=−15\cos \left( {A + B} \right) = - \frac{1}{5}.
44 Easy
44 of 147
Application +4 | -1
If cos⁡  A+cos⁡  B=m\cos \,\,A + \cos \,\,B = m and sin⁡A+sin⁡B=n\sin A + \sin B = n where m,   n≠0m,\;\,n \ne 0, then sin⁡(A+B)\sin \left( {A + B} \right) is equal to

View answer and explanation
Correct option: B
2mnm2+n2\frac{{2mn}}{{{m^2} + {n^2}}}
Option B (2mnm2+n2\frac{{2mn}}{{{m^2} + {n^2}}}) is correct because it is derived from trigonometric identities and algebraic manipulation of the given equations.
45 Easy
45 of 147
Application +4 | -1
The value of tan⁡  70∘−tan⁡  20∘tan⁡  50∘=\frac{{\tan \,\,70^\circ - \tan \,\,20^\circ }}{{\tan \,\,50^\circ }} =
View answer and explanation
Correct option: A
2
Option A is correct because it correctly simplifies the trigonometric expression using the tangent subtraction formula and other trigonometric identities.

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46 Medium
46 of 147
Application +4 | -1
If tan⁡α=kcot⁡β\tan \alpha = k\cot \beta, then cos⁡  (α−β)cos⁡  (α+β)\frac{{\cos \,\,\left( {\alpha - \beta } \right)}}{{\cos \,\,\left( {\alpha + \beta } \right)}} is equal to
View answer and explanation
Correct option: A
1+k1−k\frac{{1 + k}}{{1 - k}}
Option A (1+k1−k\frac{{1 + k}}{{1 - k}}) is correct because it is derived by using the given relation between tan⁡α\tan \alpha and tan⁡β\tan \beta and applying trigonometric identities.
47 Easy
47 of 147
Application +4 | -1
If (1+tan⁡  θ)  (1+tan⁡  ϕ)=2\left( {1 + \tan \,\,\theta } \right)\,\,\left( {1 + \tan \,\,\phi } \right) = 2, then θ+ϕ=\theta + \phi =
View answer and explanation
Correct option: B
45∘45^\circ
Option B (45∘45^\circ) is correct because it is derived from the tangent addition formula after simplifying the given equation.
48 Medium
48 of 147
Analyzing +4 | -1
If sin⁡4α+4cos⁡4β+2=42sin⁡αcos⁡β;α,β∈[0,π]\sin ^{4} \alpha+4 \cos ^{4} \beta+2=4 \sqrt{2} \sin \alpha \cos \beta ; \alpha, \beta \in[0, \pi], then cos⁡(α+β)−cos⁡(α−β)\cos (\alpha+\beta)-\cos (\alpha-\beta) is equal to:
View answer and explanation
Correct option: D
−2-\sqrt{2}
Option D (−2-\sqrt{2}) is correct because after applying AM-GM inequality and solving for α\alpha and β\beta, the expression cos⁡(α+β)−cos⁡(α−β)\cos(\alpha+\beta)-\cos(\alpha-\beta) simplifies to −2-\sqrt{2}.
49 Easy
49 of 147
Application +4 | -1
cos⁡  α  sin⁡(β−γ)+cos⁡  β  sin⁡  (γ−β)+cos⁡  γ  sin⁡(α−β)=\cos \,\,\alpha \,\,\sin \left( {\beta - \gamma } \right) + \cos \,\,\beta \,\,\sin \,\,\left( {\gamma - \beta } \right) + \cos \,\,\gamma \,\,\sin \left( {\alpha - \beta } \right) =
View answer and explanation
Correct option: A
0\mathrm{0}
Option A (0) is correct because the expression simplifies to zero after applying trigonometric identities and cancellations.
50 Medium
50 of 147
Application +4 | -1
If 3  sin⁡  θ+cos⁡  θ>0\sqrt 3 \,\,\sin \,\,\theta + \cos \,\,\theta > 0, then θ\theta lies in the interval
View answer and explanation
Correct option: B
(−π/6,   5π/6)\left( { - \pi /6,\,\;5\pi /6} \right)
Option B ((−π/6,  5π/6)\left( { - \pi /6,\;5\pi /6} \right)) is correct because it represents the interval where the given inequality holds true after applying trigonometric identities.
51 Easy
51 of 147
Analyzing +4 | -1
The expression cos⁡  10 π13+cos⁡  8 π13+cos⁡  3 π13+cos⁡  5 π13\cos \,\,\frac{{10\,\pi }}{{13}} + \cos \,\,\frac{{8\,\pi }}{{13}} + \cos \,\,\frac{{3\,\pi }}{{13}} + \cos \,\,\frac{{5\,\pi }}{{13}} is equal to
View answer and explanation
Correct option: B
0
Option B (0) is correct because the sum of the cosine terms can be paired and simplified using trigonometric identities, resulting in zero.
52 Easy
52 of 147
Application +4 | -1
If α,  β\alpha ,\,\,\beta are the solutions of a  tan⁡  θ+b  sec⁡  θ=ca\,\,\tan \,\,\theta + b\,\,\sec \,\,\theta = c, then tan⁡  (α+β)=\tan \,\,\left( {\alpha + \beta } \right) =
View answer and explanation
Correct option: A
2   aca2−c2\frac{{2\;\,ac}}{{{a^2} - {c^2}}}
Option B (2   aca2−c2\frac{{2\;\,ac}}{{{{a^2} - {c^2}}}}) is the correct answer because it is derived from the trigonometric equation and the properties of roots of a quadratic equation.
53 Medium
53 of 147
Analyzing +4 | -1
If α,β,γ∈(0,π2)\alpha, \beta, \gamma \in\left(0, \frac{\pi}{2}\right), then sin⁡α+sin⁡β+sin⁡γ\sin \alpha+\sin \beta+\sin \gamma
View answer and explanation
Correct option: A
>sin⁡(α+β+γ)> \sin (\alpha+\beta+\gamma)
Option A (>sin⁡(α+β+γ)> \sin(\alpha + \beta + \gamma)) is correct because the difference sin⁡α+sin⁡β+sin⁡γ−sin⁡(α+β+γ)\sin \alpha + \sin \beta + \sin \gamma - \sin(\alpha + \beta + \gamma) is positive given the constraints on α,β,γ\alpha, \beta, \gamma.
54 Medium
54 of 147
Application +4 | -1
If tan⁡(α−β)=sin⁡(2β)3−cos⁡(2β)\tan (\alpha-\beta)=\frac{\sin (2 \beta)}{3-\cos (2 \beta)}, then tan⁡α=f(β)\tan \alpha=f(\beta). The value of f(π3)f\left(\frac{\pi}{3}\right) equals:
View answer and explanation
Correct option: C
232 \sqrt{3}
Option C is correct because substituting β=π3\beta = \frac{{\pi}}{{3}} and simplifying leads to tan⁡α=23\tan \alpha = 2\sqrt{{3}}.
55 Easy
55 of 147
Application +4 | -1
If A+C=B{A}+{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\tan B-\tan C-\tan A) is correct because it is derived from applying the tangent addition formula to the equation A + C = B.
56 Easy
56 of 147
Application +4 | -1
tan⁡  2π5−tan⁡  π15−3  tan⁡  2π5tan⁡  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.
57 Medium
57 of 147
Application +4 | -1
If tan⁡α=mm+1\tan \alpha = \frac{m}{{m + 1}} and tan⁡β=12m+1\tan \beta = \frac{1}{{2m + 1}}, then α+β\alpha + \beta is equal to
View answer and explanation
Correct option: B
π4\frac{\pi }{4}
Option B (π4\frac{\pi}{4}) is correct because by using the formula for tan⁡(α+β)\tan(\alpha + \beta) and substituting the given values, we find that tan⁡(α+β)=1\tan(\alpha + \beta) = 1, which implies α+β=π4\alpha + \beta = \frac{\pi}{4}.
58 Easy
58 of 147
Application +4 | -1
tan⁡  25∘+tan⁡  20∘+tan⁡  25∘  tan⁡  20∘\tan \,\,25^\circ + \tan \,\,20^\circ + \tan \,\,25^\circ \,\,\tan \,\,20^\circ is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because by using the tangent addition formula and the fact that tan⁡45∘=1\tan 45^\circ = 1, the given expression simplifies to 1.
59 Medium
59 of 147
Analyzing +4 | -1
If tan⁡(α−β)=sin⁡2β3−cos⁡2β\tan (\alpha-\beta)=\frac{\sin 2 \beta}{3-\cos 2 \beta}, then tan⁡α=f(β)\tan \alpha=f(\beta). The value of f(π4)f\left(\frac{\pi}{4}\right) equals
View answer and explanation
Correct option: A
2
Option A (2) is correct because substituting β=π4\beta = \frac{\pi}{4} into the given equation and solving for tan⁡α\tan \alpha yields 2.
60 Easy
60 of 147
Application +4 | -1
If A+B+C=180∘A+B+C=180^{\circ}, then the value of tan⁡A+tan⁡B+tan⁡C\tan A+\tan B+\tan C is
View answer and explanation
Correct option: A
≥33\geq 3 \sqrt{3}
Option A (≥33\geq 3 \sqrt{3}) is correct because when A+B+C=180∘A+B+C = 180^{\circ}, tan⁡A+tan⁡B+tan⁡C=tan⁡Atan⁡Btan⁡C\tan A + \tan B + \tan C = \tan A \tan B \tan C, and by AM-GM inequality, the minimum value is 333 \sqrt{3}.
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