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JEE • MATHEMATICS • COMPOUND ANGLES

Compound angles practice · Medium

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Showing 31-51 of 51 questions
31 Medium
31 of 51
Application +4 | -1
An angle α\alpha is divided into two parts so that the ratio of the tangents of these parts is λ\lambda. If the difference between these parts is xx then sin⁡x/sin⁡α\sin x / \sin \alpha is equal to
View answer and explanation
Correct option: C
λ−1λ+1\frac{\lambda-1}{\lambda+1}
Option C, λ−1λ+1\frac{\lambda-1}{\lambda+1}, is correct because it directly follows from applying the tangent subtraction and addition formulas and then using the sine subtraction and addition formulas.
32 Medium
32 of 51
Application +4 | -1
The smallest positive values of xx and yy which satisfy tan⁡(x−y)=1,  sec⁡(x+y)=23\tan \left( {x - y} \right) = 1,\,\,\sec \left( {x + y} \right) = \frac{2}{{\sqrt 3 }} are
View answer and explanation
Correct option: A
x=25   π24,  y=7  π24x = \frac{{25\,\;\pi }}{{24}},\,\,y = \frac{{7\;\pi }}{{24}}
Option A (x=25  π24,, y=7  π24x = \frac{{25\;\pi }}{{24}},\\,\,y = \frac{{7\;\pi }}{{24}}) is correct because it provides the smallest positive values of x and y that satisfy both given trigonometric equations.
33 Medium
33 of 51
Application +4 | -1
If cot⁡αcot⁡β=2\cot \alpha \cot \beta=2, then cos⁡(α+β)cos⁡(α−β)\frac{\cos (\alpha+\beta)}{\cos (\alpha-\beta)} is equal to
View answer and explanation
Correct option: A
13\frac{1}{3}
Option A (13\frac{1}{3}) is correct because by using the given equation and applying the cosine addition and subtraction formulas, the expression simplifies to 13\frac{1}{3}.
34 Medium
34 of 51
Application +4 | -1
If cos⁡Acos⁡B=n\frac{{\cos A}}{{\cos B}} = n and sin⁡Asin⁡B=m\frac{{\sin A}}{{\sin B}} = m, then (m2−n2)sin⁡2B=\left( {{m^2} - {n^2}} \right){\sin ^2}B =
View answer and explanation
Correct option: A
1−n21 - {n^2}
Option A (1−n21 - {n^2}) is correct because it is derived by manipulating the given trigonometric ratios and applying trigonometric identities.
35 Medium
35 of 51
Application +4 | -1
If A+C=BA + C = B, then tan⁡A  tan⁡B  tan⁡C=\tan A\,\,\tan B\,\,\tan C =
View answer and explanation
Correct option: B
tan⁡  B−tan⁡  C−tan⁡  A\tan \,\,B - \tan \,\,C - \tan \,\,A
Option B (p2=q(q+2){p^2} = q\left( {q + 2} \right)) is correct because it is derived by using the relationship between roots and coefficients of a quadratic equation and trigonometric identities.
36 Medium
36 of 51
Application +4 | -1
tan⁡15∘+1tan⁡165∘+1tan⁡105∘+tan⁡195∘=2a\tan 15^{\circ}+\frac{1}{\tan 165^{\circ}}+\frac{1}{\tan 105^{\circ}}+\tan 195^{\circ}=2 a. Then the value of (a+1a)\left(a+\frac{1}{a}\right) is___________
View answer and explanation
Correct option: B
−43-\frac{4}{\sqrt{3}}
Option B (-43\frac{{4}}{{\sqrt{{3}}}}) is correct because it is the simplified value of a+1aa + \frac{{1}}{{a}} after calculating aa from the given trigonometric expression.
37 Medium
37 of 51
Analyzing +4 | -1
If tan⁡  α+tan⁡  βcot⁡  α+cot⁡  β  +  {cos⁡  (α−β)  sec⁡  (α+β)+1}−1=1\frac{{\tan \,\,\alpha + \tan \,\,\beta }}{{\cot \,\,\alpha + \cot \,\,\beta }}\,\, + \,\,{\left\{ {\cos \,\,\left( {\alpha - \beta } \right)\,\,\sec \,\,\left( {\alpha + \beta } \right) + 1} \right\}^{ - 1}} = 1, then tan⁡α tan⁡  α  tan⁡  β\tan \,\,\alpha \,\,\tan \,\,\beta is equal to
View answer and explanation
Correct option: A
1
The correct answer is A (1) because simplifying the given trigonometric expression leads to tan⁡αtan⁡β=1\tan \alpha \tan \beta = 1.
38 Medium
38 of 51
Analyzing +4 | -1
If x+y−z+xyz=0x + y - z + xyz = 0, then 2x1−x2+2y1−y2−2z1−z2\frac{{2x}}{{1 - {x^2}}} + \frac{{2y}}{{1 - {y^2}}} - \frac{{2z}}{{1 - {z^2}}} is equal to
View answer and explanation
Correct option: C
8xyz[(1−x2)(1−y2)(1−z2)]\frac{8 x y z}{\left[\left(1-x^2\right)\left(1-y^2\right)\left(1-z^2\right)\right]}
Option D (−8xyz[(1−x2)(1−y2)(1−z2)]\frac{{-8 x y z}}{{\left[(1-x^2)(1-y^2)(1-z^2)\right]}}) is the correct answer because it is derived using trigonometric substitutions and the tangent addition formula.
39 Medium
39 of 51
Application +4 | -1
If tan⁡x+tan⁡(x+π/3)+\tan x+\tan (x+\pi / 3)+ tan⁡(x+2π/3)=3\tan (x+2 \pi / 3)=3, then
View answer and explanation
Correct option: C
tan⁡3x=1\tan 3 x=1
Option C (tan⁡3x=1\tan 3x = 1) is correct because simplifying the given trigonometric equation leads to the result tan⁡3x=1\tan 3x = 1.
40 Medium
40 of 51
Application +4 | -1
The value of ∑r=15cos⁡(2r−1)π11\sum\limits_{{r=1}}^5 \cos (2 r-1) \frac{\pi}{11} is
View answer and explanation
Correct option: A
12\frac{1}{2}
Option A (12\frac{1}{2}) is correct because using the sum-to-product formula and simplifying the expression leads to this result.
41 Medium
41 of 51
Application +4 | -1
The value of 16cos⁡(2π15)⋅cos⁡(4π15)cos⁡(8π15)⋅cos⁡(16π15)16 \cos \left(\frac{2 \pi}{15}\right) \cdot \cos \left(\frac{4 \pi}{15}\right) \cos \left(\frac{8 \pi}{15}\right) \cdot \cos \left(\frac{16 \pi}{15}\right) is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the expression simplifies to 1 using trigonometric identities.
42 Medium
42 of 51
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 the sine of the sum of angles is always less than the sum of the sines of the individual angles in the given interval.
43 Medium
43 of 51
Application +4 | -1
If sin⁡2x=α−1\sin 2 x=\alpha-1 and cos⁡2x=β−1\cos 2 x=\beta-1, then the value of sec⁡2x[(cos⁡2x−sin⁡2x)−2sin⁡xcos⁡x]1+sin⁡2x\frac{\sec ^2 x\left[\left(\cos ^2 x-\sin ^2 x\right)-2 \sin x \cos x\right]}{1+\sin 2 x} is equal to
View answer and explanation
Correct option: B
2α−2β\frac{2}{\alpha}-\frac{2}{\beta}
Option C (2β−2α\frac{2}{\beta}-\frac{2}{\alpha}) is correct because the expression simplifies to 2(1α−1β)2(\frac{1}{\alpha} - \frac{1}{\beta}).
44 Medium
44 of 51
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.
45 Medium
45 of 51
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.

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46 Medium
46 of 51
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.
47 Medium
47 of 51
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).
48 Medium
48 of 51
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.
49 Medium
49 of 51
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).
50 Medium
50 of 51
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.
51 Medium
51 of 51
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.
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