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JEE • MATHEMATICS • COMPOUND ANGLES

Compound angles practice · Mixed difficulty

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Showing 1-30 of 147 questions
1 Easy
1 of 147
Application +4 | -1
If tan⁡(A+B)=p\tan \left( {A + B} \right) = p and tan⁡(A−B)=q\tan \left( {A - B} \right) = q, then the value of tan⁡  2A\tan \,\,2A is
View answer and explanation
Correct option: D
p  +  q1  −  pq\frac{{p\,\, + \,\,q}}{{1\,\, - \,\,pq}}
Option D (p+q1−pq\frac{{p\\ + \\ q}}{{1\\ - \\ pq}}) is correct because using the tangent addition formula, tan⁡2A\tan 2A can be expressed in terms of p and q as shown.
2 Medium
2 of 147
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.
3 Medium
3 of 147
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}.
4 Easy
4 of 147
Analyzing +4 | -1
If tan⁡  A=2  tan⁡  B+cot⁡  B,    then   2  tan⁡(A−B)  \tan \,\,A = 2\,\,\tan \,\,B + \cot \,\,B,\;\;\mathrm{then}\;\,2\,\,\tan \left( {A - B} \right)\mathrm{\;} is equal to

View answer and explanation
Correct option: C
cot⁡B\cot B
Option C (cot⁡B\cot B) is correct because simplifying the given expression using trigonometric identities leads to this result.
5 Easy
5 of 147
Application +4 | -1
If A,B,CA,B,C are angles of a triangle, then the minimum value of tan⁡2  A2+tan⁡2  B2+tan⁡2  C2{\tan ^2}\,\,\frac{A}{2} + {\tan ^2}\,\,\frac{B}{2} + {\tan ^2}\,\,\frac{C}{2}, is
View answer and explanation
Correct option: B
1
Option B (1) is correct because the minimum value of the given expression is 1, derived using the relationship between the angles of a triangle and AM-GM inequality.
6 Hard
6 of 147
Application +4 | -1
If A>0,   B>0A > 0,\,\;B > 0 and A+B=π3A + B = \frac{\pi }{3}, then the maximum value of tan⁡A  tan⁡B\tan A\,\,\tan B, is

View answer and explanation
Correct option: A
13\frac{1}{3}
Option A (13\frac{1}{3}) is correct because the maximum value of tan⁡Atan⁡B\tan A \tan B is achieved when A=B=π6A = B = \frac{\pi}{6}, leading to the value 13\frac{1}{3}.
7 Medium
7 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{\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.
8 Medium
8 of 147
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.
9 Medium
9 of 147
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.
10 Easy
10 of 147
Application +4 | -1
sin⁡50∘−sin⁡70∘+sin⁡10∘=\sin 50^{\circ}-\sin 70^{\circ}+\sin 10^{\circ}= ?
View answer and explanation
Correct option: B
0
Option B (0) is correct because the expression simplifies to zero using trigonometric identities.
11 Medium
11 of 147
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.
12 Easy
12 of 147
Application +4 | -1
If   cos⁡  θ=12  (x+1x),   then    12  (x2+1x2)\;\cos \,\,\mathrm{\theta } = \frac{1}{2}\,\,\left( {x + \frac{1}{x}} \right),\,\;\mathrm{then}\,\,\;\frac{1}{2}\,\,\left( {{x^2} + \frac{1}{{{x^2}}}} \right) is equal to

View answer and explanation
Correct option: B
cos⁡  2θ\cos \,\,2\mathrm{\theta }
Option B (cos⁡ 2θ\cos \,2\mathrm{\theta }) is correct because the expression simplifies to cos⁡2θ\cos 2\theta using trigonometric identities and the given equation.
13 Easy
13 of 147
Application +4 | -1
If If  tan⁡  α=(1+2−x)−1,  tan⁡  β=(1+2x+1)−1,   then    α+β  \mathrm{If}\,\,\tan \,\,\alpha = {\left( {1 + {2^{ - x}}} \right)^{ - 1}},\,\,\tan \,\,\beta = {\left( {1 + {2^{x + 1}}} \right)^{ - 1}},\;\,\mathrm{then}\;\,\,\alpha + \beta \; equals

View answer and explanation
Correct option: B
π4\frac{\pi }{4}
Option B (π4\frac{\pi }{4}) is correct because by using the tangent addition formula, tan⁡(α+β)\tan(\alpha + \beta) simplifies to 1, hence α+β=π4\alpha + \beta = \frac{\pi}{4}.
14 Easy
14 of 147
Application +4 | -1
cos⁡  9∘+sin⁡  9∘cos⁡  9∘−sin⁡  9∘\frac{{\cos \,\,9^\circ + \sin \,\,9^\circ }}{{\cos \,\,9^\circ - \sin \,\,9^\circ }} is equals to
View answer and explanation
Correct option: D
tan⁡  54∘\tan \,\,54^\circ
Option D (tan⁡ 54∘\tan \,54^\circ) is correct because it is obtained by dividing both numerator and denominator by cos⁡ 9∘\cos \,9^\circ and using the tangent addition formula.
15 Medium
15 of 147
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}}.

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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
16 Easy
16 of 147
Analyzing +4 | -1
If α\alpha and β\beta are solutions of sin⁡2x+asin⁡x+b=0\sin ^2 x+a \sin x+b=0 as well that of cos⁡2x+ccos⁡x+d=0\cos ^2 x+c \cos x+d=0, then (α+β)(\alpha+\beta) is equal to
View answer and explanation
Correct option: D
2aca2+c2\frac{2 a c}{a^2+c^2}
Option D (2aca2+c2\frac{2 a c}{a^2+c^2}) is correct because it represents sin⁡(α+β)\sin(\alpha + \beta) derived from the given conditions.
17 Medium
17 of 147
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.
18 Easy
18 of 147
Application +4 | -1
If sin⁡  A=45  \sin \,\,A = \frac{4}{5}\; and cos⁡  B=−1213\cos \,\,B = - \frac{{12}}{{13}}, where A  A\; and B  B\; lie in first and third quadrant respectively, then cos (A  +  B)\mathrm{cos}\,(A\; + \;B) is equal to
View answer and explanation
Correct option: D
−1665- \frac{{16}}{{65}}
Option D (−1665- \frac{{16}}{{65}}) is correct because it is the result of applying the cosine addition formula and using the given sine and cosine values with correct signs based on the quadrants.
19 Easy
19 of 147
Application +4 | -1
If α\alpha and β\beta be between 0\mathrm{0} and π2\frac{\pi }{2} and if cos⁡  (α+β)=1213\cos \,\,\left( {\alpha + \beta } \right) = \frac{{12}}{{13}} and sin⁡(α−β)=35\sin \left( {\alpha - \beta } \right) = \frac{3}{5}, then sin⁡  2α\sin \,\,2\alpha is equal to
View answer and explanation
Correct option: B
56/6556/65
Option B (56/65) is correct because using the given information and the formula for sin⁡(A+B)\sin(A+B), we can find sin⁡2α\sin 2\alpha.
20 Easy
20 of 147
Application +4 | -1
If   tan⁡  α=mm+1  and  tan⁡  β=12m+1,  then    α+β  \mathrm{\;}\tan \,\,\mathrm{\alpha } = \frac{\mathrm{m}}{{\mathrm{m} + 1}}\,\,\mathrm{and}\,\,\tan \,\,\mathrm{\beta } = \frac{1}{{2\mathrm{m} + 1}},\mathrm{\;then\;}\,\,\mathrm{\alpha } + \mathrm{\beta \;} is equal to

View answer and explanation
Correct option: B
π4\frac{\pi }{4}
Option B (nπ2+π8\frac{{n\pi }}{2} + \frac{\pi }{8}) is correct because by simplifying the equation using trigonometric identities, the general solution is found to be nπ2+π8\frac{{n\pi }}{2} + \frac{\pi }{8}.
21 Easy
21 of 147
Application +4 | -1
If sin⁡  (α+β)=1,  sin⁡  (α−β)=1  /2 ;   α,   β∈[0,   π/2]\sin \,\,\left( {\alpha + \beta } \right) = 1,\,\,\sin \,\,\left( {\alpha - \beta } \right) = 1\;/2\,;\,\;\alpha ,\,\;\beta \in \left[ {0,\,\;\pi /2} \right], then tan⁡  (α+2β)  tan⁡  (2α+β)\tan \,\,\left( {\alpha + 2\beta } \right)\,\,\tan \,\,\left( {2\alpha + \beta } \right) is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because the expression simplifies to 1 after finding the values of α\alpha and β\beta from the given equations and substituting them into the expression.
22 Easy
22 of 147
Application +4 | -1
If tan⁡  θ=12\tan \,\,\theta = \frac{1}{2} and tan⁡  ϕ=13\tan \,\,\phi = \frac{1}{3}, then the value of θ+ϕ\theta + \phi, is
View answer and explanation
Correct option: D
π/4\pi /4
Option D (π/4\pi /4) is correct because it results from applying the tangent addition formula and simplifying.
23 Easy
23 of 147
Application +4 | -1
tan⁡  10∘+tan⁡  35∘+tan⁡  10∘  tan⁡  35∘\tan \,\,10^\circ + \tan \,\,35^\circ + \tan \,\,10^\circ \,\,\tan \,\,35^\circ is equal to
View answer and explanation
Correct option: D
1\mathrm{1}
Option D (1) is correct because the expression simplifies to 1 using the tangent addition formula.
24 Medium
24 of 147
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.
25 Medium
25 of 147
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.
26 Medium
26 of 147
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.
27 Medium
27 of 147
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}.
28 Easy
28 of 147
Application +4 | -1
Let ABAB and CC be the angles of a plain triangle and tan⁡  A2=13,  tan⁡B2=23\tan \,\,\frac{A}{2} = \frac{1}{3},\,\,\tan \frac{B}{2} = \frac{2}{3}. Then, tan⁡C2\tan \frac{C}{2} is equal to
View answer and explanation
Correct option: A
7/97/9
Option A (7/9) is correct because it is derived by 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.
29 Easy
29 of 147
Application +4 | -1
If y=(1+tan⁡A)(1−tan⁡B)y=(1+\tan A)(1-\tan B) where A−B=π4A-B=\frac{\pi}{4} then (y+1)y+1(y+1)^{y+1} is equal to:
View answer and explanation
Correct option: C
27
Option C (27) is correct because by using the given condition A−B=π4A-B = \frac{\pi}{4}, we can find that y=2y=2, and then calculate (y+1)y+1=27(y+1)^{y+1} = 27.
30 Easy
30 of 147
Application +4 | -1
If y=(1+tan⁡A)  (1−tan⁡B)y = \left( {1 + \tan A} \right)\,\,\left( {1 - \tan B} \right), where A−B=π4A - B = \frac{\pi }{4}, then (y+1)y+1{\left( {y + 1} \right)^{y + 1}} is equal to
View answer and explanation
Correct option: C
27
Option C (27) is correct because after substituting A−B=π4A - B = \frac{\pi}{{4}} into the given equation, we find y=2y = 2, and then calculate (y+1)y+1=33=27(y + 1)^{y + 1} = 3^3 = 27.
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