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JEE • MATHEMATICS • COMPOUND ANGLES

Compound angles practice · Easy

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Practice format and chapter quiz

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Showing 1-30 of 91 questions
1 Easy
1 of 91
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 Easy
2 of 91
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.
3 Easy
3 of 91
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.
4 Easy
4 of 91
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.
5 Easy
5 of 91
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.
6 Easy
6 of 91
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}.
7 Easy
7 of 91
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.
8 Easy
8 of 91
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.
9 Easy
9 of 91
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.
10 Easy
10 of 91
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.
11 Easy
11 of 91
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}.
12 Easy
12 of 91
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.
13 Easy
13 of 91
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.
14 Easy
14 of 91
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.
15 Easy
15 of 91
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.

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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 91
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.
17 Easy
17 of 91
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.
18 Easy
18 of 91
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.
19 Easy
19 of 91
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.
20 Easy
20 of 91
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.
21 Easy
21 of 91
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.
22 Easy
22 of 91
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.
23 Easy
23 of 91
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.
24 Easy
24 of 91
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}.
25 Easy
25 of 91
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.
26 Easy
26 of 91
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.
27 Easy
27 of 91
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.
28 Easy
28 of 91
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.
29 Easy
29 of 91
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.
30 Easy
30 of 91
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.
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