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JEE • MATHEMATICS • TRIGONOMETRY RATIOS

Trigonometry ratios practice · Easy

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Practice format and chapter quiz

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 181-210 of 218 questions
181 Easy
181 of 218
Application +4 | -1
The value of 2cos⁡20∘cos⁡40∘cos⁡80∘2 \cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ} is equal to ___.
View answer and explanation
Correct option: A
0.250.25
Option A (0.25) is correct because the expression simplifies to 1/4 using the double angle formula for sine repeatedly.
182 Easy
182 of 218
Understanding +4 | -1
Assertion: The measure of rotation of a given ray about its initial point is called an angle.

Reason: The point of rotation is called a vertex.
View answer and explanation
Correct option: B
Assertion is correct, reason is correct; reason is not a correct explanation for assertion
Assertion is correct, reason is correct; reason is not a correct explanation for assertion because while both statements are true, the reason doesn't explain why the measure of rotation is called an angle.
183 Easy
183 of 218
Analyzing +4 | -1
The angle θ\theta whose cosine equals to its tangent is given by
View answer and explanation
Correct option: C
sin⁡  θ=2  sin⁡  18∘\sin \,\,\theta = 2\,\,\sin \,\,18^\circ
Option C (sin⁡ θ=2 sin⁡ 18∘\sin \,\theta = 2\,\sin \,18^\circ) is correct because by setting cos⁡θ=tan⁡θ\cos \theta = \tan \theta and solving for sin⁡θ\sin \theta, we find that sin⁡θ\sin \theta is equal to 2sin⁡18∘2\sin 18^\circ.
184 Easy
184 of 218
Application +4 | -1
In △ABC,cos⁡2A+cos⁡2B−cos⁡2C=1\triangle A B C, \cos ^2 A+\cos ^2 B-\cos ^2 C=1, then
View answer and explanation
Correct option: C
∠C=π2\angle C=\frac{\pi}{2}
Option C (∠C=π2\angle C=\frac{\pi}{2}) is correct because the given equation can be rearranged to show that a2+b2=c2a^2 + b^2 = c^2, which implies that the triangle is right-angled at C.
185 Easy
185 of 218
Application +4 | -1
If tan⁡A=(1−cos⁡B)sin⁡B\tan \mathrm{A}=\frac{(1-\cos \mathrm{B})}{\sin \mathrm{B}}, then tan⁡2 A=……\tan 2 \mathrm{~A}=\ldots \ldots
View answer and explanation
Correct option: A
tan⁡B\tan \mathrm{B}
Option A (tan⁡B\tan B) is correct because simplifying the given expression for tan⁡A\tan A and then finding tan⁡2A\tan 2A using the double angle formula results in tan⁡B\tan B.
186 Easy
186 of 218
Application +4 | -1
If cos⁡  x−24  sin⁡x=  λcos⁡  (x+α),   0<α<π2\cos \,\,x - 24\,\,\sin x = \;\lambda \cos \,\,\left( {x + \alpha } \right),\,\;0 < \alpha < \frac{\pi }{2}, be true for all x  ∈Rx\; \in R, then

View answer and explanation
Correct option: A
λ=25\lambda = 25
Option A (λ=25\lambda = 25) is correct because by rewriting the given expression and comparing coefficients, we can find that λ=25\lambda = 25.
187 Easy
187 of 218
Analyzing +4 | -1
For −π2<θ<π2,  sin⁡θ+sin⁡2θ1+cos⁡θ+cos⁡2θ- \frac{\pi }{2} < \theta < \frac{\pi }{2},\,\,\frac{{\sin \theta + \sin 2\theta }}{{1 + \cos \theta + \cos 2\theta }} lies in the interval
View answer and explanation
Correct option: A
(−∞,   ∞)\left( { - \infty ,\;\,\infty } \right)
Option A is correct because the given expression simplifies to tan θ, and for -π/2 < θ < π/2, tan θ can take any real value.
188 Easy
188 of 218
Application +4 | -1
Suppose α,β>0\alpha, \beta>0 and α+2β=π/2\alpha+2 \beta=\pi / 2, then tan⁡(α+β)−2tan⁡α−tan⁡β\tan (\alpha+\beta)-2 \tan \alpha-\tan \beta is equal to:
View answer and explanation
Correct option: A
0
Option A (0) is correct because using the given condition α+2β=π2\alpha + 2\beta = \frac{\pi}{2}, we can show that tan⁡(α+β)−2tan⁡α−tan⁡β=0\tan (\alpha+\beta)-2 \tan \alpha-\tan \beta = 0.
189 Easy
189 of 218
Application +4 | -1
If cos⁡(α+β)=4/5\cos (\alpha+\beta)=4 / 5 and sin⁡(α−β)=5/13\sin (\alpha-\beta)=5 / 13 where 0≤α,β≤π/40 \leq \alpha, \beta \leq \pi / 4, then tan⁡2α=\tan 2 \alpha=
View answer and explanation
Correct option: D
56/3356 / 33
Option D (56/3356 / 33) is correct because it is the result of applying the tangent addition formula after finding tan⁡(α+β)\tan(\alpha+\beta) and tan⁡(α−β)\tan(\alpha-\beta).
190 Easy
190 of 218
Application +4 | -1
If cos⁡θ=−35\cos \theta=\frac{-3}{5} and π<θ<3π2\pi < \theta < \frac{3 \pi}{2}, then the value of (cosec⁡θ+cot⁡θsec⁡θ−tan⁡θ)\left(\frac{\operatorname{cosec} \theta+\cot \theta}{\sec \theta-\tan \theta}\right) is equal to 1m\frac{1}{m}. Value of mm is
View answer and explanation
Correct option: A
6
Option A (1) is correct because simplifying the given expression using trigonometric identities results in 1.
191 Easy
191 of 218
Application +4 | -1
If θ=π/8\theta=\pi / 8, and f(θ)=cos⁡6θ+cos⁡63θ+f(\theta)=\cos ^6 \theta+\cos ^6 3 \theta+ cos⁡65θ+cos⁡67θ\cos ^6 5 \theta+\cos ^6 7 \theta then f(θ)f(\theta) is equal to:
View answer and explanation
Correct option: C
5/45 / 4
The correct answer is C (5/4) because by using trigonometric identities and substituting the value of θ, we can simplify the expression to obtain the result.
192 Easy
192 of 218
Application +4 | -1
The sum of the interior angles of a regular polygon is twice the sum of its exterior angles, the polygon is a
View answer and explanation
Correct option: A
Hexagon
Option A (Hexagon) is correct because the sum of interior angles being twice the sum of exterior angles leads to the equation (n−2)×180∘=2×360∘(n-2) \times 180^{\circ} = 2 \times 360^{\circ}, which solves to n=6.
193 Easy
193 of 218
Application +4 | -1
If π<θ<2π\pi < \theta < 2\pi, then 1+cos⁡  θ1−cos⁡  θ\sqrt {\frac{{1 + \cos \,\,\theta }}{{1 - \cos \,\,\theta }}} is equal to
View answer and explanation
Correct option: D
−cosec   θ−cot⁡  θ- \mathrm{cosec}\,\;\theta - \cot \,\,\theta
Option D (−cosec ;θ−cot⁡θ- \mathrm{cosec}\,\\;\theta - \cot \theta) is correct because, given π<θ<2π\pi < \theta < 2\pi, sin⁡θ\sin \theta is negative, which affects the sign when simplifying the square root.
194 Easy
194 of 218
Application +4 | -1
sin⁡70∘+cos⁡40∘cos⁡70∘+sin⁡40∘=\frac{\sin 70^{\circ}+\cos 40^{\circ}}{\cos 70^{\circ}+\sin 40^{\circ}}=
View answer and explanation
Correct option: C
3\sqrt{3}
Option C (3\sqrt{3}) is correct because after converting cosine to sine and applying sum-to-product formulas, the expression simplifies to 3\sqrt{3}.
195 Easy
195 of 218
Application +4 | -1
If sin⁡θ+sin⁡2θ+sin⁡3θ=sin⁡α\sin \theta+\sin 2 \theta+\sin 3 \theta=\sin \alpha and cos⁡θ\cos \theta +cos⁡2θ+cos⁡3θ=cos⁡α+\cos 2 \theta+\cos 3 \theta=\cos \alpha, then θ\theta is equal to
View answer and explanation
Correct option: A
α/2\alpha / 2
Option A (α/2\alpha / 2) is correct because dividing the sine equation by the cosine equation results in tan⁡2θ=tan⁡α\tan 2\theta = \tan \alpha, which implies θ=α/2\theta = \alpha / 2.

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196 Easy
196 of 218
Application +4 | -1
The radius of the circle whose arc of length 15 cm15 \mathrm{~cm} makes an angle of 3/43 / 4 radian at the centre is
View answer and explanation
Correct option: B
20 cm20 \mathrm{~cm}
Option B (20 cm20 \mathrm{~cm}) is correct because it is derived from the formula relating arc length, radius, and angle in radians: ℓ=rθ\ell = r\theta.
197 Easy
197 of 218
Application +4 | -1
If tan⁡θ=17\tan \theta=\frac{1}{\sqrt{7}}, then (cosec⁡2θ−sec⁡2θcosec⁡2θ+sec⁡2θ)\left(\frac{\operatorname{cosec}^2 \theta-\sec ^2 \theta}{\operatorname{cosec}^2 \theta+\sec ^2 \theta}\right) is equal to mm+1\frac{m}{m+1}. The value of mm is
View answer and explanation
Correct option: A
3
Option A (1) is correct because evaluating the given expression using the values of trigonometric functions at the specified angles results in 1.
198 Easy
198 of 218
Application +4 | -1
If sin⁡  θ+cosec    θ=2\sin \,\,\theta + \mathrm{cosec}\,\,\;\theta = 2, the value of sin⁡10  θ+cosec10  θ{\sin ^{10}}\,\,\theta + \mathrm{cose}{\mathrm{c}^{10}}\,\,\theta is
View answer and explanation
Correct option: A
2
Option A is correct because when sin θ + cosec θ = 2, it implies sin θ = 1, and thus sin¹⁰ θ + cosec¹⁰ θ = 1 + 1 = 2.
199 Easy
199 of 218
Understanding +4 | -1
The value of the expression cos  1∘  .  cos⁡  2∘…..cos⁡  179∘\mathrm{cos}\,\,1^\circ \,\,.\,\,\cos \,\,2^\circ \ldots ..\cos \,\,179^\circ equals
View answer and explanation
Correct option: A
0\mathrm{0}
Option A (0) is correct because the product includes cos(90°), which is 0, making the entire product 0.
200 Easy
200 of 218
Application +4 | -1
The value of cot⁡54∘tan⁡36∘+tan⁡20∘cot⁡70∘\frac{\cot 54^{\circ}}{\tan 36^{\circ}}+\frac{\tan 20^{\circ}}{\cot 70^{\circ}} is
View answer and explanation
Correct option: B
2
Option B (2) is correct because cot⁡54∘=tan⁡36∘\cot 54^{\circ} = \tan 36^{\circ} and tan⁡20∘=cot⁡70∘\tan 20^{\circ} = \cot 70^{\circ}, so the expression simplifies to 1 + 1 = 2.
201 Easy
201 of 218
Application +4 | -1
The product (cos⁡x2)⋅(cos⁡x4)⋅(cos⁡x8)⋯⋯⋅(cos⁡x256)\left(\cos \frac{x}{2}\right) \cdot\left(\cos \frac{x}{4}\right) \cdot\left(\cos \frac{x}{8}\right) \cdots \cdots \cdot\left(\cos \frac{x}{256}\right) is equal to:
View answer and explanation
Correct option: B
sin⁡x256sin⁡x256\frac{\sin x}{256 \sin \frac{x}{256}}
Option B (sin⁡x256sin⁡x256\frac{{\sin x}}{{256 \sin \frac{{x}}{{256}}}}) is correct because repeatedly multiplying and dividing by sine terms and using the identity sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2 \sin \theta \cos \theta leads to the given expression.
202 Easy
202 of 218
Application +4 | -1
If tan⁡A−tan⁡B=12\tan A-\tan B=\frac{1}{2} and cot⁡A−cot⁡B=13\cot A-\cot B=\frac{1}{3}, then cot⁡(A−B)\cot (A-B) is equal to:
View answer and explanation
Correct option: B
-1
The correct answer is B (-1) because by manipulating the given equations and using the cotangent subtraction formula, we find the value of cot(A-B).
203 Easy
203 of 218
Application +4 | -1
The expression 3  {sin⁡4(3 π2−α)+sin⁡4  (3 π−α)}−2  {sin⁡6(π2+α)+sin⁡6  (5 π−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\mathrm{1}
Option B (1) is correct because after simplifying the trigonometric expressions and applying trigonometric identities, the expression evaluates to 1.
204 Easy
204 of 218
Application +4 | -1
If cosec⁡θ=p+qp−q\operatorname{cosec} \theta=\frac{p+q}{p-q}, then cot⁡(π4+θ2)=\cot \left(\frac{\pi}{4}+\frac{\theta}{2}\right)=
View answer and explanation
Correct option: B
qp\sqrt{\frac{q}{p}}
Option B (qp\sqrt{\frac{{q}}{{p}}}) is correct because after applying componendo and dividendo and simplifying, cot⁡(π4+θ2)=qp\cot(\frac{{\pi}}{{4}} + \frac{{\theta}}{{2}}) = \sqrt{\frac{{q}}{{p}}}.
205 Easy
205 of 218
Application +4 | -1
Let f(x)=17−sin⁡5xf(x)=\frac{1}{7-\sin 5 x} be a function defined on RR. Then the range of the function f(x)f(x) is equal to:
View answer and explanation
Correct option: D
[18,16]\left[\frac{1}{8}, \frac{1}{6}\right]
Option D ([18,16]\left[\frac{{1}}{{8}}, \frac{{1}}{{6}}\right]) is correct because it represents the range of the function, determined by considering the minimum and maximum values of the sine function.
206 Easy
206 of 218
Application +4 | -1
If sin⁡(x−y)=cos⁡(x+y)=12\sin \left( {x - y} \right) = \cos \left( {x + y} \right) = \frac{1}{2}, the values of x\mathrm{x} and y\mathrm{y} 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 these values satisfy both given equations sin⁡(x−y)=1/2\sin(x-y) = 1/2 and cos⁡(x+y)=1/2\cos(x+y) = 1/2.
207 Easy
207 of 218
Application +4 | -1
If sin⁡Asin⁡B=32\frac{\sin A}{\sin B}=\frac{\sqrt{3}}{2} and cos⁡Acos⁡B=52\frac{\cos A}{\cos B}=\frac{\sqrt{5}}{2}, 0<A,B<π/20 < A, B < \pi / 2, then tan⁡A+tan⁡B\tan A+\tan B is equal to
View answer and explanation
Correct option: D
(3+5)/5(\sqrt{3}+\sqrt{5}) / \sqrt{5}
Option C (136,116\frac{{13}}{{6}}, \frac{{11}}{{6}}) is correct because it satisfies both equations in the system.
208 Easy
208 of 218
Understanding +4 | -1
In a triangle ABCA B C the value of sin⁡A+sin⁡B+sin⁡C\sin A+\sin B+\sin C is
View answer and explanation
Correct option: B
4cos⁡A2cos⁡B2cos⁡C24 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}
Option B (4cos⁡A2cos⁡B2cos⁡C24 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}) is correct because it is the result of applying trigonometric identities to simplify the expression sin⁡A+sin⁡B+sin⁡C\sin A + \sin B + \sin C.
209 Easy
209 of 218
Application +4 | -1
tan⁡2θ+cot⁡2θ1\tan ^2 \theta+\cot ^2 \theta{ }_1. Is
View answer and explanation
Correct option: A
≥2\geq 2
Option A (≥2\geq 2) is correct because by using the AM-GM inequality or by considering the square of the difference, it can be shown that tan⁡2θ+cot⁡2θ≥2\tan^2 \theta + \cot^2 \theta \geq 2.
210 Easy
210 of 218
Application +4 | -1
cos⁡10∘+sin⁡10∘cos⁡10∘−sin⁡10∘\frac{\cos 10^{\circ}+\sin 10^{\circ}}{\cos 10^{\circ}-\sin 10^{\circ}} is equal to
View answer and explanation
Correct option: A
tan⁡55∘\tan 55^{\circ}
Option A (tan⁡55∘\tan 55^{\circ}) is correct because the expression simplifies to tan⁡(45∘+10∘)\tan(45^{\circ} + 10^{\circ}) using the tangent addition formula.
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