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JEE • MATHEMATICS • COMPOUND ANGLES

Compound angles practice · Medium

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Showing 1-30 of 51 questions
1 Medium
1 of 51
Analyzing +4 | -1
The value of cos⁡  x  cos⁡  y  sin⁡  (x−y)+cos⁡  y  cos⁡  z  sin⁡  (y−z)\cos \,\,x\,\,\cos \,\,y\,\,\sin \,\,\left( {x - y} \right) + \cos \,\,y\,\,\cos \,\,z\,\,\sin \,\,\left( {y - z} \right)

+cos⁡  z  cos⁡  x  sin⁡  (z−x)+sin⁡  (x−y)sin⁡  (y−z)  sin⁡  (z−x)+ \cos \,\,z\,\,\cos \,\,x\,\,\sin \,\,\left( {z - x} \right) + \sin \,\,\left( {x - y} \right)\sin \,\,\left( {y - z} \right)\,\,\sin \,\,\left( {z - x} \right), is
View answer and explanation
Correct option: A
0\mathrm{0}
Option A (0) is correct because the entire expression simplifies to zero after applying trigonometric identities and summing the terms.
2 Medium
2 of 51
Application +4 | -1
The sides of a triangle are sin⁡α,cos⁡α\sin \alpha, \cos \alpha and 1+sin⁡αcos⁡α\sqrt{1+\sin \alpha \cos \alpha} for some 0<α<π20<\alpha<\frac{\pi}{2}. Then, the greatest angle of the triangle is
View answer and explanation
Correct option: C
120∘120^{\circ}
Option C (120∘120^{\circ}) is correct because by using the Law of Cosines, we can find the largest angle of the triangle, which is 120∘120^{\circ}.
3 Medium
3 of 51
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{\alpha }{2} + \tan \,\,\frac{\beta }{2} + \tan \,\,\frac{\gamma }{2} = \tan \,\,\frac{\alpha }{2}\,\,\tan \,\,\frac{\beta }{2}\,\,\tan \,\,\frac{\gamma }{2}
Option A (tan⁡ α2+tan⁡ β2+tan⁡ γ2=tan⁡ α2tan⁡ β2tan⁡ γ2\tan \,\frac{\alpha}{2} + \tan \,\frac{\beta}{2} + \tan \,\frac{\gamma}{2} = \tan \,\frac{\alpha}{2}\tan \,\frac{\beta}{2}\tan \,\frac{\gamma}{2}) is correct because it is the result of applying the tangent addition formula with the given condition α+β+γ=2π\alpha + \beta + \gamma = 2\pi.
4 Medium
4 of 51
Application +4 | -1
sin⁡π9sin⁡2π9sin⁡4π9sin⁡π3\sin \frac{\pi}{9} \sin \frac{2 \pi}{9} \sin \frac{4 \pi}{9} \sin \frac{\pi}{3} is equal to
View answer and explanation
Correct option: C
316\frac{3}{16}
Option C (316\frac{3}{16}) is correct because it is the result of applying the identity sin⁡θsin⁡(60∘−θ)sin⁡(60∘+θ)=14sin⁡3θ\sin \theta \sin (60^{\circ} - \theta) \sin (60^{\circ} + \theta) = \frac{1}{4} \sin 3\theta and simplifying.
5 Medium
5 of 51
Application +4 | -1
tan⁡A+2tan⁡2A+4tan⁡4A+8cot⁡8A=\tan A + 2\tan 2A + 4\tan 4A + 8\cot 8A =
View answer and explanation
Correct option: A
cot⁡A\cot A
Option A (cot⁡A\cot A) is correct because the given expression simplifies to cot⁡A\cot A by repeatedly using the identity cot⁡x−tan⁡x=2cot⁡2x\cot x - \tan x = 2 \cot 2x.
6 Medium
6 of 51
Application +4 | -1
If sin⁡θ=3sin⁡(θ+2α),\sin \theta = 3\sin \left( {\theta + 2\alpha } \right), then the value of tan⁡(θ+α)+2tan⁡α\tan \left( {\theta + \alpha } \right) + 2\tan \alpha is
View answer and explanation
Correct option: A
0
Option A (0) is correct because by using trigonometric identities and simplification, the expression tan⁡(θ+α)+2tan⁡α\tan (\theta + \alpha) + 2\tan \alpha simplifies to 0.
7 Medium
7 of 51
Application +4 | -1
If cos⁡25∘+sin⁡25∘=k\cos 25^{\circ}+\sin 25^{\circ}=k, then cos⁡20∘\cos 20^{\circ} is equal to
View answer and explanation
Correct option: A
k2\frac{k}{\sqrt{2}}
Option A (k2\frac{k}{\sqrt{2}}) is correct because by using the cosine subtraction formula and the given equation, cos⁡20∘\cos 20^{\circ} is found to be k2\frac{k}{\sqrt{2}}.
8 Medium
8 of 51
Application +4 | -1
If tan⁡β=nsin⁡αcos⁡α1−nsin⁡2α\tan \beta=\frac{n \sin \alpha \cos \alpha}{1-n \sin ^2 \alpha}, then tan⁡(α−β)\tan (\alpha-\beta) is equal to
View answer and explanation
Correct option: B
(1−n)tan⁡α(1-n) \tan \alpha
Option B ((1−n)tan⁡α(1-n) \tan \alpha) is correct because by using the tangent subtraction formula, tan⁡(α−β)\tan(\alpha - \beta) simplifies to (1−n)tan⁡α(1-n)\tan \alpha.
9 Medium
9 of 51
Application +4 | -1
The value of cos⁡12∘+cos⁡84∘+cos⁡156∘+cos⁡132∘\cos 12^{\circ}+\cos 84^{\circ}+\cos 156^{\circ}+\cos 132^{\circ} is
View answer and explanation
Correct option: D
−12\frac{-1}{2}
Option D is correct because by applying the sum-to-product trigonometric identities and simplifying, the expression evaluates to -1/2.
10 Medium
10 of 51
Application +4 | -1
If tan⁡  α=(1+2−x)−1,  tan⁡  β=(1+2x+1)−1\tan \,\,\alpha = {\left( {1 + {2^{ - x}}} \right)^{ - 1}},\,\,\tan \,\,\beta = {\left( {1 + {2^{x + 1}}} \right)^{ - 1}}, then α+β\alpha + \beta equals
View answer and explanation
Correct option: B
π/4\pi /4
Option B (π/4\pi /4) is correct because the sum α+β\alpha + \beta simplifies to π/4\pi/4 after applying the tangent addition formula and given conditions.
11 Medium
11 of 51
Analyzing +4 | -1
tan⁡  5x  tan⁡  3x  tan⁡  2x=\tan \,\,5x\,\,\tan \,\,3x\,\,\tan \,\,2x =
View answer and explanation
Correct option: A
tan⁡  5x−tan⁡  3x−tan⁡  2x\tan \,\,5x - \tan \,\,3x - \tan \,\,2x
Option A (tan⁡5x−tan⁡3x−tan⁡2x\tan 5x - \tan 3x - \tan 2x) is correct because it is derived from applying the tangent addition formula and rearranging the terms.
12 Medium
12 of 51
Application +4 | -1
Let f(θ)=cot⁡θ1+cot⁡θf(\theta)=\frac{\cot \theta}{1+\cot \theta} and α+β=5π4\alpha+\beta=\frac{5 \pi}{4}. Then the value of f(α)⋅f(β)f(\alpha) \cdot f(\beta) is
View answer and explanation
Correct option: C
12\frac{1}{2}
Option C (12\frac{1}{2}) is correct because by using the given equation and simplifying the expression, the value of f(α)⋅f(β)f(\alpha) \cdot f(\beta) is found to be 12\frac{1}{2}.
13 Medium
13 of 51
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.
14 Medium
14 of 51
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.
15 Medium
15 of 51
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.

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16 Medium
16 of 51
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.
17 Medium
17 of 51
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.
18 Medium
18 of 51
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

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.
19 Medium
19 of 51
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.
20 Medium
20 of 51
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}.
21 Medium
21 of 51
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.
22 Medium
22 of 51
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.
23 Medium
23 of 51
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}}.
24 Medium
24 of 51
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}.
25 Medium
25 of 51
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.
26 Medium
26 of 51
Application +4 | -1
tan⁡5π12−tan⁡π12−3tan⁡5π12tan⁡π12\tan \frac{5 \pi}{12}-\tan \frac{\pi}{12}-\sqrt{3} \tan \frac{5 \pi}{12} \tan \frac{\pi}{12} is equal to
View answer and explanation
Correct option: D
3\sqrt{3}
Option D (3\sqrt{3}) is correct because using the tangent subtraction formula, the expression simplifies to tan⁡(5π12−π12)=tan⁡(π3)=3\tan(\frac{5\pi}{12} - \frac{\pi}{12}) = \tan(\frac{\pi}{3}) = \sqrt{3}.
27 Medium
27 of 51
Application +4 | -1
If acos⁡3θ+3acos⁡θsin⁡2θ=xa \cos ^3 \theta+3 a \cos \theta \sin ^2 \theta=x and asin⁡3θ+3acos⁡2θsin⁡θ=ya \sin ^3 \theta+3 a \cos ^2 \theta \sin \theta=y, then (x+y)2/3+(x−y)2/3=(x+y)^{2 / 3}+(x-y)^{2 / 3}=
View answer and explanation
Correct option: A
2a2/32 a^{2 / 3}
Option A (2a2/32 a^{2 / 3}) is correct because simplifying the given equations and using the binomial theorem leads to this result.
28 Medium
28 of 51
Application +4 | -1
If CC is an obtuse angle in tringle, then
View answer and explanation
Correct option: A
tan⁡Atan⁡B<1\tan A\tan B < 1
The correct answer is A (tan⁡Atan⁡B<1\tan A\tan B < 1) because if C is obtuse, then A+B is acute, and the tangent of an acute angle is positive.
29 Medium
29 of 51
Analyzing +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) =
View answer and explanation
Correct option: A
a2+  b2{a^2} + \,\,{b^2}
Option A (a2+  b2{a^2} + \,\,{b^2}) is correct because the trigonometric identity and given equations simplify to this expression.
30 Medium
30 of 51
Application +4 | -1
If cos⁡  θ=817\cos \,\,\theta = \frac{8}{{17}} and θ\theta lies in the first quadrant, then the value of cos⁡  (30∘+θ)+cos⁡  (45∘−θ)+cos⁡  (120∘−θ)\cos \,\,\left( {30^\circ + \theta } \right) + \cos \,\,\left( {45^\circ - \theta } \right) + \cos \,\,\left( {120^\circ - \theta } \right), is
View answer and explanation
Correct option: A
2317  (3−12+12)\frac{{23}}{{17}}\,\,\left( {\frac{{\sqrt 3 - 1}}{2} + \frac{1}{{\sqrt 2 }}} \right)
Option A is correct because it correctly uses trigonometric identities and the given cosine value to simplify the expression and find the final value.
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