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JEE • MATHEMATICS • TRIGONOMETRY RATIOS

Trigonometry ratios practice · Medium

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 121-150 of 157 questions
121 Medium
121 of 157
Application +4 | -1
In a △ABC,cos⁡A=35\triangle A B C, \cos A=\frac{3}{5} and cos⁡B=513\cos B=\frac{5}{13}. The value of cos⁡C\cos C can be
View answer and explanation
Correct option: C
3365\frac{33}{65}
Option C (3365\frac{33}{65}) is correct because it is the value of cos C that satisfies the trigonometric relationships within a triangle, given the values of cos A and cos B.
122 Medium
122 of 157
Application +4 | -1
If 3π4<a<π\frac{3 \pi}{4} < a < \pi, then cosec⁡2α+2cot⁡α=\sqrt{\operatorname{cosec}^2 \alpha+2 \cot \alpha}=
View answer and explanation
Correct option: C
−1−cot⁡α-1-\cot \alpha
Option C (−1−cot⁡α-1-\cot \alpha) is correct because within the given range for α\alpha, 1+cot⁡α1 + \cot \alpha is negative, so the square root evaluates to the negative of that expression.
123 Medium
123 of 157
Analyzing +4 | -1
The greatest value of sin⁡3x+cos⁡3x{\sin ^3}x + {\cos ^3}x is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the maximum value of the function is 1, which can be found by analyzing its derivative and critical points.
124 Medium
124 of 157
Application +4 | -1
Sides of a △ABC\triangle A B C are in AP, if a<a< minimum {b,c}\{b, c\}, then cos⁡A\cos A is equal to
View answer and explanation
Correct option: D
4c−3b2c\frac{4 c-3 b}{2 c}
Option D (4c−3b2c\frac{4 c-3 b}{2 c}) is correct because by considering the possible arithmetic progressions and applying the cosine rule, this expression for cos⁡A\cos A is obtained.
125 Medium
125 of 157
Application +4 | -1
If acos⁡3α+3acos⁡αsin⁡2α=ma \cos ^3 \alpha+3 a \cos \alpha \sin ^2 \alpha=m and asin⁡3α+3acos⁡2αsin⁡α=na \sin ^3 \alpha+3 a \cos ^2 \alpha \sin \alpha=n, then (m+n)2/3+(m+n)^{2 / 3}+ (m−n)2/3(m-n)^{2 / 3} is equal to
View answer and explanation
Correct option: C
2a2/32 a^{2 / 3}
Option C (2a2/3a^{2/3}) is correct because, given the equations for m and n, we can find (m+n)(2/3) + (m-n)(2/3) by using the binomial expansion and trigonometric identities, which simplifies to 2a2/3a^{2/3}.
126 Medium
126 of 157
Application +4 | -1
If tan⁡θ+sin⁡θ=m\tan \theta+\sin \theta=m and tan⁡θ−sin⁡θ=n\tan \theta-\sin \theta=n, then
View answer and explanation
Correct option: D
m2−n2=4mnm^2-n^2=4 \sqrt{m n}
Option D (m2−n2=4mnm^2-n^2=4 \sqrt{m n}) is correct because by using the given equations, we can derive the relationship between m, n, tan θ, and sin θ.
127 Medium
127 of 157
Application +4 | -1
If 2cos⁡A=x+1/x2cos⁡B=y+1/y2 \cos A=x+1 / x 2 \cos B=y+1 / y then find the value of &#39; kk &#39; such that kcos⁡(A−B)=xy+yxk \cos (A-B)=\frac{x}{y}+\frac{y}{x}.
View answer and explanation
Correct option: A
2
Option A (2) is correct because by expressing x and y in terms of cosine and sine functions and substituting into the given expression, we can simplify it to 2cos⁡(A−B)2\cos(A-B), thus k=2.
128 Medium
128 of 157
Application +4 | -1
The sides of a △ABC\triangle A B C are in AP, if the angles AA and CC are the greatest and the smallest angles respectively, then 4(1−cos⁡A)(1−cos⁡C)4(1-\cos A)(1-\cos C) is equal to
View answer and explanation
Correct option: B
4cos⁡2A+C24 \cos ^2 \frac{A+C}{2}
Option B (4cos⁡2A+C24 \cos ^2 \frac{{A+C}}{{2}}) is correct because by using the properties of AP and trigonometric identities, the expression simplifies to this form.
129 Medium
129 of 157
Understanding +4 | -1
cos⁡4θ−sin⁡4θ\cos ^4 \theta-\sin ^4 \theta is equal to
View answer and explanation
Correct option: B
2cos⁡2θ−12 \cos ^2 \theta-1
Option B (2cos⁡2θ−12 \cos ^2 \theta-1) is correct because cos⁡4θ−sin⁡4θ\cos ^4 \theta - \sin ^4 \theta simplifies to cos⁡2θ\cos 2\theta, which is equal to 2cos⁡2θ−12 \cos ^2 \theta - 1.
130 Medium
130 of 157
Analyzing +4 | -1
3  (sin⁡x−cos⁡x)4+6(sin⁡x+cos⁡x)2+4(sin⁡6x+cos⁡6x)3\;{\left( {\sin x - \cos x} \right)^4} + 6{\left( {\sin x + \cos x} \right)^2} + 4\left( {{{\sin }^6}x + {{\cos }^6}x} \right) is equal to
View answer and explanation
Correct option: B
13
Option B (13) is correct because by expanding and simplifying the given trigonometric expression, we find that it equals 13.
131 Medium
131 of 157
Application +4 | -1
cos⁡2A2+cos⁡2B2+cos⁡2C2=\cos ^2 \frac{A}{2}+\cos ^2 \frac{B}{2}+\cos ^2 \frac{C}{2}=
View answer and explanation
Correct option: B
2+2sin⁡A2sin⁡B2sin⁡C22+2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}
Option B (2+2sin⁡A2sin⁡B2sin⁡C22+2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}) is correct because by applying half-angle formulas and trigonometric identities, the expression simplifies to the given form.
132 Medium
132 of 157
Application +4 | -1
If α+β−γ=π\alpha+\beta-\gamma=\pi, then sin⁡2α+sin⁡2β−sin⁡2γ=\sin ^2 \alpha+\sin ^2 \beta-\sin ^2 \gamma=
View answer and explanation
Correct option: C
2sin⁡αsin⁡βcos⁡γ2 \sin \alpha \sin \beta \cos \gamma
Option A (2sin⁡αsin⁡βcos⁡γ2 \sin \alpha \sin \beta \cos \gamma) is correct because by using trigonometric identities and the given condition α+β−γ=π\alpha + \beta - \gamma = \pi, the expression simplifies to 2sin⁡αsin⁡βcos⁡γ2 \sin \alpha \sin \beta \cos \gamma.
133 Medium
133 of 157
Application +4 | -1
If sin⁡(x+y)sin⁡(x−y)=a+ba−b\frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}, then tan⁡xtan⁡y\frac{\tan x}{\tan y} is equal to
View answer and explanation
Correct option: B
ab\frac{a}{b}
Option B (ab\frac{{a}}{{b}}) is correct because by applying the componendo and dividendo rule and simplifying, the ratio tan⁡xtan⁡y\frac{{\tan x}}{{\tan y}} is found to be ab\frac{{a}}{{b}}.
134 Medium
134 of 157
Application +4 | -1
The value of ∑k=13  cos⁡2(2k−1)π12\sum \limits_{k = 1}^3 \,\,{\cos ^2}\left( {2k - 1} \right)\frac{\pi }{{12}}, is
View answer and explanation
Correct option: D
3/23/2
Option D (3/2) is correct because by substituting the values of k and using trigonometric identities, the summation simplifies to 32\frac{3}{2}.
135 Medium
135 of 157
Application +4 | -1
If x+y−z+xyz=0x+y-z+x y z=0, then 2x1−x2+2y1−y2−2z1−z2\frac{2 x}{1-x^2}+\frac{2 y}{1-y^2}-\frac{2 z}{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 C (8xyz[(1−x2)(1−y2)(1−z2)]\frac{{8 x y z}}{{\left[(1-x^2)(1-y^2)(1-z^2)\right]}}) is correct because by substituting x=tan⁡Ax = \tan A, y=tan⁡By = \tan B, and −z=tan⁡C-z = \tan C and using trigonometric identities, the expression simplifies to the given form.

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136 Medium
136 of 157
Application +4 | -1
The numerical value of sin⁡π18⋅sin⁡5π18⋅sin⁡7π18\sin \frac{\pi}{18} \cdot \sin \frac{5 \pi}{18} \cdot \sin \frac{7 \pi}{18} is equal to
View answer and explanation
Correct option: B
1/81 / 8
Option B (1/81 / 8) is correct because the given expression can be simplified using the identity sin⁡x⋅sin⁡(60∘−x)⋅sin⁡(60∘+x)=14sin⁡3x\sin x \cdot \sin (60^\circ - x) \cdot \sin (60^\circ + x) = \frac{{1}}{{4}} \sin 3x, leading to the solution 1/81/8.
137 Medium
137 of 157
Application +4 | -1
If cos⁡α=2cos⁡β−12−cos⁡β(0<α<β<π)\cos \alpha=\frac{2 \cos \beta-1}{2-\cos \beta}(0 < \alpha < \beta < \pi), then tan⁡α/2tan⁡β/2\frac{\tan \alpha / 2}{\tan \beta / 2} is equal to
View answer and explanation
Correct option: C
3\sqrt{3}
Option C (3\sqrt{3}) is correct because after applying trigonometric identities and simplifying, the ratio tan⁡α2tan⁡β2\frac{{\tan \frac{{\alpha}}{{2}}}}{{\tan \frac{{\beta}}{{2}}}} equals 3\sqrt{3}.
138 Medium
138 of 157
Application +4 | -1
Find the value of cot⁡5∘cot⁡10∘…..cot⁡85∘\cot 5^{\circ} \cot 10^{\circ} \ldots . . \cot 85^{\circ}.
View answer and explanation
Correct option: A
1
Option A (1) is correct because pairing the cotangents with their complementary angles results in a product of 1.
139 Medium
139 of 157
Application +4 | -1
If α+β+γ=π\alpha+\beta+\gamma=\pi then the minimum value of cos⁡A+cos⁡B+cos⁡C\cos A+\cos B+\cos C
View answer and explanation
Correct option: D
is -3
Option D (is -3) is correct because the minimum value of cosine is -1, and if A=B=C=π, then cosA+cosB+cosC = -3.
140 Medium
140 of 157
Analyzing +4 | -1
The expression (1+tan⁡  x+tan⁡2x) (1−cot⁡x+cot⁡2x)(1 + \tan \,\,x + {\tan ^2}x)\,(1 - \cot x + {\cot ^2}x) has the positive value for xx given by
View answer and explanation
Correct option: C
For all x∈R
Option C (For all x∈R) is correct because the given expression simplifies to a form that is always positive for all real values of x.
141 Medium
141 of 157
Application +4 | -1
If x=sec⁡θ−tan⁡θ,y=cosec⁡θ+cot⁡θx=\sec \theta-\tan \theta, y=\operatorname{cosec} \theta+\cot \theta, then xy+1=….x y+1=\ldots ..
View answer and explanation
Correct option: C
y−xy-x
Option C (y−xy-x) is correct because simplifying the expression xy+1xy+1 using trigonometric identities results in y−xy-x.
142 Medium
142 of 157
Analyzing +4 | -1
If 0<x<π2,0 < x < \frac{\pi }{2},, then
View answer and explanation
Correct option: A
cos⁡(sin⁡x)>cos⁡x\cos \left( {\sin x} \right) > \cos x
Option A (cos⁡(sin⁡x)>cos⁡x\cos(\sin x) > \cos x) is correct because cos⁡x\cos x is a decreasing function on (0,π2)(0, \frac{\pi}{2}) and sin⁡x<x\sin x < x for 0<x<π20 < x < \frac{\pi}{2}.
143 Medium
143 of 157
Application +4 | -1
If sin⁡(x−y)=cos⁡(x+y)=1/2\sin (x-y)=\cos (x+y)=1 / 2 then xx and yy lying between 0∘0^{\circ} and 90∘90^{\circ} are given by -
View answer and explanation
Correct option: D
x=45∘,y=15∘x=45^{\circ}, y=15^{\circ}
Option D (x=45∘,y=15∘x=45^{\circ}, y=15^{\circ}) is correct because solving the system of equations sin⁡(x−y)=1/2\sin (x-y) = 1/2 and cos⁡(x+y)=1/2\cos (x+y) = 1/2 yields these values for x and y within the given range.
144 Medium
144 of 157
Application +4 | -1
In any triangle ABC,  ∑  sin⁡2A+sin⁡A+1sin⁡AABC,\,\,\sum \,\,\frac{{{{\sin }^2}A + \sin A + 1}}{{\sin A}} is always greater than
View answer and explanation
Correct option: A
9
Option A (9) is correct because each term in the summation is greater than or equal to 3, and there are three terms in the summation, resulting in a value greater than or equal to 9.
145 Medium
145 of 157
Application +4 | -1
In a quadrilateral, the measure of the angles are in A.P. If the ratio of the measures of the greatest and the smallest angles is 2:12: 1, then the measure of the greatest angle in radians is ...... .
View answer and explanation
Correct option: B
2π3\frac{2 \pi}{3}
Option B ([−1,12]\left[-1, \frac{{1}}{{\sqrt{{2}}}}\right]) is correct because solving the inequality sin⁡−1x≤cos⁡−1x\sin^{-1}x \le \cos^{-1}x leads to this interval.
146 Medium
146 of 157
Application +4 | -1
If p=2sin⁡θ1+cos⁡θ+sin⁡θp=\frac{2 \sin \theta}{1+\cos \theta+\sin \theta}, and q=cos⁡θ1+sin⁡θq=\frac{\cos \theta}{1+\sin \theta}, then
View answer and explanation
Correct option: D
q+p=1q+p=1
Option D (q+p=1q+p=1) is correct because the given expressions for pp and qq can be simplified using trigonometric identities to show that their sum equals 1.
147 Medium
147 of 157
Application +4 | -1
In a △ABC, a2sin⁡2B+b2sin⁡2A\triangle {ABC}, \, {a}^2 \sin 2 {B}+{b}^2 \sin 2 {A} is equal to (where Δ\Delta is area of △ABC\triangle {ABC})
View answer and explanation
Correct option: A
4Δ4 \Delta
Option A (4Δ4 \Delta) is correct because a2sin⁡2B+b2sin⁡2Aa^2 \sin 2B + b^2 \sin 2A simplifies to 4Δ4 \Delta using trigonometric identities and the area formula.
148 Medium
148 of 157
Analyzing +4 | -1
sin⁡2  θ=4 xy(x+y)2{\sin ^2}\,\,\theta = \frac{{4\,xy}}{{{{\left( {x + y} \right)}^2}}} is true if and only if
View answer and explanation
Correct option: B
x=y,  x≠0,   y≠0x = y,\;x \ne 0,\;\,y \ne 0
Option B (x=y,  x≠0,   y≠0x = y,\;x \ne 0,\;\,y \ne 0) is correct because the equation holds true only when x and y are equal and non-zero, ensuring that the sine squared value is less than or equal to 1.
149 Medium
149 of 157
Analyzing +4 | -1
The value of the expression 1−sin⁡2  y1+cos⁡  y+1+cos⁡  ysin⁡  y−sin⁡  y1−cos⁡  y1 - \frac{{{{\sin }^2}\,\,y}}{{1 + \cos \,\,y}} + \frac{{1 + \cos \,\,y}}{{\sin \,\,y}} - \frac{{\sin \,\,y}}{{1 - \cos \,\,y}} is equal to
View answer and explanation
Correct option: D
cos⁡  y\cos \,\,y
Option D (cos y) is correct because the expression simplifies to cos y after applying trigonometric identities and algebraic manipulation.
150 Medium
150 of 157
Application +4 | -1
If   tan⁡  A+sin⁡  A=m    and  tan⁡  A−sin⁡  A=n  \;\tan \,\,A + \sin \,\,A = m\,\,\;\mathrm{and}\,\,\tan \,\,A - \sin \,\,A = n\; then (m2−n2)2mn\frac{{{{\left( {{m^2} - {n^2}} \right)}^2}}}{{mn}} is equal to
View answer and explanation
Correct option: C
16
Option C (16) is correct because it is the simplified value of the expression after substituting the given equations and using trigonometric identities.
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