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JEE • MATHEMATICS • TRIGONOMETRY RATIOS

Trigonometry ratios practice · Mixed difficulty

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Showing 181-210 of 394 questions

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181 Easy
181 of 394
Application +4 | -1
If sin⁡θ+cosec⁡θ=2\sin \theta+\operatorname{cosec} \theta=2, the value of sin⁡10θ+cosec⁡10θ\sin ^{10} \theta+\operatorname{cosec}^{10} \theta is
View answer and explanation
Correct option: A
2
Option A (2) is correct because if sin⁡θ+cosec⁡θ=2\sin \theta + \operatorname{cosec} \theta = 2, then sin⁡θ=1\sin \theta = 1, and thus sin⁡10θ+cosec⁡10θ=110+110=2\sin^{10} \theta + \operatorname{cosec}^{10} \theta = 1^{10} + 1^{10} = 2.
182 Medium
182 of 394
Analyzing +4 | -1
If sin⁡  x+sin⁡2x=1\sin \,\,x + {\sin ^2}x = 1, then the value of cos⁡12  x+3  cos⁡10x+3  cos⁡8 x+cos⁡6  x+2  cos⁡4  x+cos⁡2x−2{\cos ^{12}}\,\,x + 3\,\,{\cos ^{10}}x + 3\,\,{\cos ^8}\,x + {\cos ^6}\,\,x + 2\,\,{\cos ^4}\,\,x + {\cos ^2}x - 2, is equal to
View answer and explanation
Correct option: D
sin⁡2  x{\sin ^2}\,\,x
Option D (sin⁡2  x{\sin ^2}\,\,x) is correct because the given expression simplifies to sin⁡2  x{\sin ^2}\,\,x using the trigonometric identity and the given condition.
183 Medium
183 of 394
Application +4 | -1
If xcos⁡θ+ysin⁡θ=zx \cos \theta+y \sin \theta=z, then what is the value of (xsin⁡θ−ycos⁡θ)2(x \sin \theta-y \cos \theta)^{2} ?
View answer and explanation
Correct option: A
x2+y2−z2x^{2}+y^{2}-z^{2}
Option A (x2+y2−z2x^{2}+y^{2}-z^{2}) is correct because it is derived by manipulating the given equation and using trigonometric identities.
184 Medium
184 of 394
Application +4 | -1
If cot⁡A=tan⁡((n−1)A),\cot A = \tan ((n - 1)A), where n∈N,n \in N, then a value of A can be
View answer and explanation
Correct option: A
π/2n\pi / 2 n
Option A (π/2n\pi / 2 n) is correct because substituting this value into the given equation satisfies the trigonometric relationship between cotangent and tangent.
185 Hard
185 of 394
Application +4 | -1
The minimum value of 27cos⁡2x.81sin⁡2x{27^{\cos 2x}}{.81^{\sin 2x}}, is
View answer and explanation
Correct option: A
1243\frac{1}{{243}}
Option A (1243\frac{1}{{243}}) is correct because the minimum value is found by minimizing the exponent of 3 using trigonometric identities.
186 Medium
186 of 394
Application +4 | -1
The value of 3  [sin⁡4(3π2−α)+sin⁡4(3π+α)]−2[sin⁡6(π2+α)+sin⁡6(5π−α)]  3\,\,\left[ {{{\sin }^4}\left( {\frac{{3\pi }}{2} - \alpha } \right) + {{\sin }^4}\left( {3\pi + \alpha } \right)} \right] - 2\left[ {{{\sin }^6}\left( {\frac{\pi }{2} + \alpha } \right) + {{\sin }^6}\left( {5\pi - \alpha } \right)} \right]\; is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because after simplifying the trigonometric expressions and applying trigonometric identities, the expression evaluates to 1.
187 Medium
187 of 394
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
1/21 / 2
Option C (1/21/2) is correct because by using the given condition α+β=5π4\alpha + \beta = \frac{{5\pi}}{{4}} and trigonometric identities, f(α)⋅f(β)f(\alpha) \cdot f(\beta) simplifies to 1/21/2.
188 Easy
188 of 394
Analyzing +4 | -1
All the pairs (x, y) that satisfy the inequality 2sin⁡2x−2sin⁡x+5.14sin⁡2y≤1{2^{\sqrt {{{\sin }^2}x - 2\sin x + 5} }}.\frac{1}{{{4^{{{\sin }^2}y}}}} \le 1 also satisfy the equation
View answer and explanation
Correct option: B
sin⁡x=∣sin⁡y∣\sin x = \left| {\sin y} \right|
Option B (sin⁡x=∣sin⁡y∣\sin x = \left| {\sin y} \right|) is correct because the inequality holds only when sin⁡x=1\sin x = 1 and sin⁡2y=1\sin^2 y = 1, which implies sin⁡x=∣sin⁡y∣\sin x = |\sin y|.
189 Easy
189 of 394
Application +4 | -1
The value of cosec⁡(−1410)∘\operatorname{cosec}(-1410)^{\circ} is equal to
View answer and explanation
Correct option: A
2
Option A (2) is correct because \operatorname{cosec}(-1410)° simplifies to \operatorname{cosec}(30)°, which equals 2.
190 Medium
190 of 394
Application +4 | -1
sin⁡2A+sin⁡2B−sin⁡2C=\sin ^2 A+\sin ^2 B-\sin ^2 C=
View answer and explanation
Correct option: D
2sin⁡Asin⁡Bcos⁡C2 \sin A \sin B \cos C
Option D (2sin⁡Asin⁡Bcos⁡C2 \sin A \sin B \cos C) is correct because by using trigonometric identities and sum-to-product formulas, the expression simplifies to the given form.
191 Medium
191 of 394
Application +4 | -1
The new position of the point (1,2)(1,2) under rotation through an angle of 90∘90^{\circ} about the origin in anticlockwise direction is
View answer and explanation
Correct option: A
(−2,1)(-2,1)
Option A ((-2,1)) is correct because it represents the new coordinates of the point (1,2) after a 90-degree anticlockwise rotation about the origin.
192 Hard
192 of 394
Application +4 | -1
If x=sin⁡θ∣sin⁡θ∣,y=cos⁡θ∣cos⁡θ∣x=\sin \theta|\sin \theta|, y=\cos \theta|\cos \theta|, where 99π2≤θ≤50π\frac{99 \pi}{2} \leq \theta \leq 50 \pi, then
View answer and explanation
Correct option: D
y−x=1y-x=1
Option D (y−x=1y-x=1) is correct because within the specified range of θ\theta, x becomes −sin⁡2θ-\sin^2 \theta and y becomes cos⁡2θ\cos^2 \theta, leading to y−x=cos⁡2θ+sin⁡2θ=1y-x = \cos^2 \theta + \sin^2 \theta = 1.
193 Easy
193 of 394
Application +4 | -1
16sin⁡(20∘)sin⁡(40∘)sin⁡(80∘)16 \sin \left(20^{\circ}\right) \sin \left(40^{\circ}\right) \sin \left(80^{\circ}\right) is equal to:
View answer and explanation
Correct option: B
232 \sqrt{3}
Option B (232 \sqrt{{3}}) is correct because it uses the identity sin⁡θ⋅sin⁡(60∘−θ)⋅sin⁡(60∘+θ)=14sin⁡3θ\sin \theta \cdot \sin (60^\circ - \theta) \cdot \sin (60^\circ + \theta) = \frac{{1}}{{4}} \sin 3\theta to simplify the expression.
194 Hard
194 of 394
Application +4 | -1
If ABCDABCD is a convex quadrilateral such that 4sec⁡A+5=04\sec A + 5 = 0 then the quadratic equation whose roots are tan⁡A\tan A and cosec  A\mathrm{cosec}\;A is
View answer and explanation
Correct option: B
12x2−11x−15=012{x^2} - 11x - 15 = 0
The correct answer is 12x2−11x−15=012{x^2} - 11x - 15 = 0 because it is the quadratic equation formed with roots tan⁡A\tan A and csc⁡A\csc A derived from the given condition.
195 Easy
195 of 394
Analyzing +4 | -1
If x  cos⁡  α+ysin⁡  α=2a,   x  cos⁡  α+y  sin⁡  β=2ax\,\,\cos \,\,\alpha + y\sin \,\,\alpha = 2a,\;\,x\,\,\cos \,\,\alpha + y\,\,\sin \,\,\beta = 2a and 2sin⁡  α2  sin⁡  β2=12\sin \,\,\frac{\alpha }{2}\,\,\sin \,\,\frac{\beta }{2} = 1, then

View answer and explanation
Correct option: C
y2=4a  (α−x){y^2} = 4a\,\,\left( {\alpha - x} \right)
Option C (y2=4a (α−x){y^2} = 4a\,\left( {\alpha - x} \right)) is correct because it is derived from the given trigonometric equations and the condition 2sin⁡ α2 sin⁡ β2=12\sin \,\frac{\alpha }{2}\,\sin \,\frac{\beta }{2} = 1.

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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
196 Easy
196 of 394
Application +4 | -1
If tan⁡A2=32\tan \frac{A}{2}=\frac{3}{2}, then 1+cos⁡A1−cos⁡A=\frac{1+\cos A}{1-\cos A}=
View answer and explanation
Correct option: D
4/94 / 9
Option D (49\frac{{4}}{{9}}) is correct because 1+cos⁡A1−cos⁡A\frac{{1 + \cos A}}{{1 - \cos A}} is equal to cot⁡2(A2)\cot^2(\frac{{A}}{{2}}), and since tan⁡(A2)=32\tan(\frac{{A}}{{2}}) = \frac{{3}}{{2}}, then cot⁡2(A2)=(23)2=49\cot^2(\frac{{A}}{{2}}) = (\frac{{2}}{{3}})^2 = \frac{{4}}{{9}}.
197 Easy
197 of 394
Application +4 | -1
The maximum value of 1+8  sin⁡2x2  cos⁡2x21 + 8\,\,{\sin ^2}{x^2}\,\,{\cos ^2}{x^2}, is
View answer and explanation
Correct option: A
3\mathrm{3}
Option A is correct because the maximum value is found by simplifying the expression using trigonometric identities and considering the range of the cosine function.
198 Easy
198 of 394
Analyzing +4 | -1
The maximum value of cos⁡  x{cos⁡  x1−sin⁡  x  +  1−sin⁡  xcos⁡  x}\cos \,\,x\left\{ {\frac{{\cos \,\,x}}{{1 - \sin \,\,x}}\,\, + \,\,\frac{{1 - \sin \,\,x}}{{\cos \,\,x}}} \right\}, is
View answer and explanation
Correct option: C
2
Option C (2) is correct because the given expression simplifies to 2 for all valid values of x.
199 Medium
199 of 394
Application +4 | -1
The value of tan⁡A+tan⁡(60∘+A)−tan⁡(60∘−A)\tan \mathrm{A}+\tan \left(60^{\circ}+\mathrm{A}\right)-\tan \left(60^{\circ}-\mathrm{A}\right) is
View answer and explanation
Correct option: C
3tan⁡3A3 \tan 3 \mathrm{A}
Option C (3tan⁡3A3 \tan 3 \mathrm{A}) is correct because by using the tangent addition and subtraction formulas, the expression can be simplified to 3tan⁡3A3 \tan 3A.
200 Medium
200 of 394
Application +4 | -1
The value of 2(sin⁡6θ+cos⁡6θ)−3(sin⁡4θ+cos⁡4θ)+12\left(\sin ^6 \theta+\cos ^6 \theta\right)-3\left(\sin ^4 \theta+\cos ^4 \theta\right)+1 is
View answer and explanation
Correct option: B
0
Option B (0) is correct because by using trigonometric identities, the expression 2(sin⁡6θ+cos⁡6θ)−3(sin⁡4θ+cos⁡4θ)+12(\sin^6 \theta + \cos^6 \theta) - 3(\sin^4 \theta + \cos^4 \theta) + 1 simplifies to 0.
201 Easy
201 of 394
Application +4 | -1
cos⁡  1∘+  cos⁡  2∘+cos⁡  3∘+…+cos⁡  180∘=\cos \,\,1^\circ + \,\,\cos \,\,2^\circ + \cos \,\,3^\circ + \ldots + \cos \,\,180^\circ =
View answer and explanation
Correct option: D
−1- 1
Option D (−1-1) is correct because the sum of the cosine series from 1° to 180° simplifies to -1 due to symmetry and the value of cos(180°).
202 Hard
202 of 394
Application +4 | -1
A watch which gains 5 seconds in 3 minutes was set right at 7am7 \mathrm{am}. In the afternoon of the same day, when the watch indicated quarter past 4 o'clock, the true time was
View answer and explanation
Correct option: B
14⋅2237min⁡14 \cdot \frac{22}{37} \min to 4
Option B (14223714 \frac{22}{37} min to 4) is correct because it calculates the true time by accounting for the watch's gain over the period from 7 am to 4:15 pm.
203 Easy
203 of 394
Analyzing +4 | -1
If x=∑n=0∞(−1)ntan⁡2nθx = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}{{\tan }^{2n}}\theta } and y=∑n=0∞cos⁡2nθy = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\theta } for 0<θ<π40 < \theta < \frac{\pi }{4} then
View answer and explanation
Correct option: B
y(1−x)=1y(1 - x) = 1
Option B (y(1−x)=1y(1 - x) = 1) is correct because by simplifying the given infinite sums for x and y, we can derive the relationship y(1−x)=1y(1-x) = 1.
204 Medium
204 of 394
Application +4 | -1
If α,β,γ∈[0,π]\alpha, \beta, \gamma \in[0, \pi] and α,β,γ\alpha, \beta, \gamma are in AP, then sin⁡α−sin⁡γcos⁡γ−cos⁡α\frac{\sin \alpha-\sin \gamma}{\cos \gamma-\cos \alpha} is equal to
View answer and explanation
Correct option: C
cot⁡β\cot \beta
Option C (cot⁡β\cot \beta) is correct because, given that α,β,γ\alpha, \beta, \gamma are in AP, we can simplify the expression sin⁡α−sin⁡γcos⁡γ−cos⁡α\frac{{\sin \alpha - \sin \gamma}}{{\cos \gamma - \cos \alpha}} to cot⁡β\cot \beta.
205 Medium
205 of 394
Application +4 | -1
If sin⁡α+sin⁡β=a\sin \alpha+\sin \beta=a and cos⁡α−cos⁡β=b\cos \alpha-\cos \beta=b, then tan⁡α−β2\tan \frac{\alpha-\beta}{2} is equal to
View answer and explanation
Correct option: B
−b/a-b / a
Option B (−b/a-b / a) is correct because by using sum-to-product trigonometric identities and dividing the resulting equations, we can find that tan⁡α−β2=−ba\tan \frac{{\alpha-\beta}}{{2}} = -\frac{{b}}{{a}}.
206 Easy
206 of 394
Application +4 | -1
If sin  +cosec = 2\mathrm{sin\; + cosec}\,\mathrm{ = }\,\mathrm{2}, then sin⁡2  θ+cosec2 θ{\sin ^2}\,\,\mathrm{\theta } + \mathrm{cose}{\mathrm{c}^2}\,\mathrm{\theta } is equal to
View answer and explanation
Correct option: C
2\mathrm{2}
Option C (2) is correct because squaring the given equation and using the reciprocal identity, we find that sin⁡2 θ+cosec2 θ=2{\sin ^2}\,\theta + \mathrm{cose}{\mathrm{c}^2}\,\theta = 2.
207 Easy
207 of 394
Application +4 | -1
Let f(θ)=(1+cos⁡θ)(1+cos⁡3θ)f(\theta)=(1+\cos \theta)(1+\cos 3 \theta) (1+cos⁡5θ)(1+cos⁡7θ)(1+\cos 5 \theta)(1+\cos 7 \theta) then f(−π8)f\left(-\frac{\pi}{8}\right) is equal to:
View answer and explanation
Correct option: D
1/81 / 8
The correct answer is D (1/8) because by using the property f(-θ) = f(θ) and trigonometric identities, we can simplify the expression to obtain the result.
208 Medium
208 of 394
Application +4 | -1
If 0<x<π2,0 < x < \frac{\pi }{2}, then the minimum value of cos⁡3xsin⁡x+sin⁡3xcos⁡x\frac{{{{\cos }^3}x}}{{\sin x}} + \frac{{{{\sin }^3}x}}{{\cos x}} is
View answer and explanation
Correct option: D
1
Option D (1) is correct because the minimum value is found using AM ≥ GM inequality, simplifying to sin⁡2x\sin 2x, and then recognizing that the minimum value of sin⁡2x\sin 2x in the given interval is 1 when x=π/4x = \pi/4.
209 Easy
209 of 394
Application +4 | -1
If tan⁡  θ=17\tan \,\,\theta = \frac{1}{{\sqrt 7 }}, then (cosec2 θ−sec⁡2θ)(cosec2 θ+sec⁡2θ)\frac{{\left( {\mathrm{cose}{\mathrm{c}^2}\,\theta - {{\sec }^2}\theta } \right)}}{{(\mathrm{cose}{\mathrm{c}^2}\,\theta + {{\sec }^2}\theta )}} is equal to
View answer and explanation
Correct option: B
34\frac{3}{4}
Option B (34\frac{3}{4}) is correct because it is obtained by substituting the given value of tan⁡θ\tan \theta and simplifying the expression using trigonometric identities.
210 Easy
210 of 394
Analyzing +4 | -1
If L=sin⁡2(π16)−sin⁡2(π8)L = {\sin ^2}\left( {\frac{\pi }{{16}}} \right) - {\sin ^2}\left( {\frac{\pi }{8}} \right) and M=cos⁡2(π16)−sin⁡2(π8), M = {\cos ^2}\left( {\frac{\pi }{{16}}} \right) - {\sin ^2}\left( {\frac{\pi }{8}} \right),\, then
View answer and explanation
Correct option: D
M=122+12cos⁡π8M = \frac{1}{{2\sqrt 2 }} + \frac{1}{2}\cos \frac{\pi }{8}
Option D (M=122+12cos⁡π8M = \frac{1}{{2\sqrt 2 }} + \frac{1}{2}\cos \frac{\pi }{8}) is correct because it follows from applying trigonometric identities to simplify the expression for M.
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