Skip to main content
Back to TRIGONOMETRY RATIOS

Your Progress

0 / 394 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • MATHEMATICS • TRIGONOMETRY RATIOS

Trigonometry ratios practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

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.

Take a chapter quiz →
Showing 121-150 of 394 questions

Preparing practice and saved progress…

121 Medium
121 of 394
Application +4 | -1
Given that tan⁡A\tan A and tan⁡B\tan B are the roots of x2−px+x^2-p x+ q=0q=0, then the value of sin⁡2(A+B)\sin ^2(A+B) is
View answer and explanation
Correct option: A
p2{p2+(1−q)2}\frac{p^2}{\left\{p^2+(1-q)^2\right\}}
Option A (p2{p2+(1−q)2}\frac{p^2}{\{p^2+(1-q)^2\}}) is correct because by using the relationship between the roots and coefficients of the quadratic equation and the tangent addition formula, the expression simplifies to the given form.
122 Medium
122 of 394
Understanding +4 | -1
(1−sin⁡A+cos⁡A)2(1-\sin A+\cos A)^{2} is equal to
View answer and explanation
Correct option: B
2(1−sin⁡A)(1+cos⁡A)2(1-\sin A)(1+\cos A)
Option B is correct because expanding and simplifying the expression leads to 2(1 - sin A)(1 + cos A).
123 Easy
123 of 394
Application +4 | -1
If A+B+C=(2n+1)πA+B+C=(2 n+1) \pi, then tan⁡A+tan⁡B+tan⁡C\tan A+\tan B+\tan C is equal to
View answer and explanation
Correct option: A
tan⁡Atan⁡Btan⁡C\tan A \tan B \tan C
Option A (tan⁡Atan⁡Btan⁡C\tan A \tan B \tan C) is correct because when A+B+C=(2n+1)πA+B+C=(2n+1)\pi, the identity tan⁡A+tan⁡B+tan⁡C=tan⁡Atan⁡Btan⁡C\tan A + \tan B + \tan C = \tan A \tan B \tan C holds true.
124 Easy
124 of 394
Application +4 | -1
In a △ABC\triangle A B C, if a,ba, b and AA are given, then there are two triangles with third sides c1c_1 and c2c_2, then (c1−c2)\left(c_1-c_2\right) is
View answer and explanation
Correct option: C
2a2−b2sin⁡2A2 \sqrt{a^2-b^2 \sin ^2 A}
Option C (2a2−b2sin⁡2A2 \sqrt{a^2-b^2 \sin ^2 A}) is correct because it is derived from solving the quadratic equation for the third side cc using the Law of Cosines and then finding the difference between the two possible solutions.
125 Medium
125 of 394
Application +4 | -1
If   cot  θ+tan⁡  θ=m    and  sec⁡  θ−cos⁡  θ=n\mathrm{\;cot}\,\,\mathrm{\theta } + \tan \,\,\mathrm{\theta } = m\mathrm{\;}\,\,\mathrm{and}\,\,\sec \,\,\mathrm{\theta } - \cos \,\,\mathrm{\theta } = n then which of the following is correct ?

View answer and explanation
Correct option: A
m  (mn2)1/3−n  (nm2)1/3=1m\,\,{\left( {m{n^2}} \right)^{1/3}} - n\,\,{\left( {n{m^2}} \right)^{1/3}} = 1
Option A (m  (mn2)1/3−n  (nm2)1/3=1m\,\,{\left( {m{n^2}} \right)^{1/3}} - n\,\,{\left( {n{m^2}} \right)^{1/3}} = 1) is correct because it is the simplified form of the equation after substituting and using trigonometric identities.
126 Medium
126 of 394
Application +4 | -1
If sin⁡  A−6  cos⁡  A=7  cos⁡  A\sin \,\,A - \sqrt 6 \,\,\cos \,\,A = \sqrt 7 \,\,\cos \,\,A, then cos⁡  A+6  sin⁡  A\cos \,\,A + \sqrt 6 \,\,\sin \,\,A is equal to
View answer and explanation
Correct option: B
− 7 sin⁡  A- \,\sqrt 7 \,\sin \,\,A
Option B (− 7 sin⁡ A- \,\sqrt 7 \,\sin \,A) is correct because it is derived by substituting A with 90° + A in the given equation.
127 Easy
127 of 394
Application +4 | -1
In any △ABC,4△(cot⁡A+cot⁡B+cot⁡C)\triangle A B C, 4 \triangle(\cot A+\cot B+\cot C) is equal to
View answer and explanation
Correct option: C
a2+b2+c2a^2+b^2+c^2
Option C (a2+b2+c2a^2+b^2+c^2) is correct because the expression 4△(cot⁡A+cot⁡B+cot⁡C)4 \triangle(\cot A+\cot B+\cot C) simplifies to a2+b2+c2a^2+b^2+c^2 using trigonometric identities and area formulas.
128 Easy
128 of 394
Application +4 | -1
The value of

cos⁡  (270∘+θ)  cos⁡  (90∘−θ)−sin⁡  (270∘−θ)  cos⁡  θ\cos \,\,\left( {270^\circ + \theta } \right)\,\,\cos \,\,\left( {90^\circ - \theta } \right) - \sin \,\,\left( {270^\circ - \theta } \right)\,\,\cos \,\,\theta is
View answer and explanation
Correct option: D
1\mathrm{1}
Option D is correct because simplifying the trigonometric expression using trigonometric identities results in 1.
129 Medium
129 of 394
Application +4 | -1
If sin⁡x+sin⁡2x−1\sin x+\sin ^2 x-1, then cos⁡14x+3cos⁡10x+3sin⁡4x+\cos ^{14} x+3 \cos ^{10} x+3 \sin ^4 x+ cos⁡4x\cos ^4 x is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because the given condition leads to sin⁡x=cos⁡2x\sin x = \cos^2 x, and substituting this into the expression simplifies it to 1.
130 Easy
130 of 394
Application +4 | -1
cos⁡11∘−sin⁡11∘cos⁡11∘+sin⁡11∘\frac{\cos 11^{\circ}-\sin 11^{\circ}}{\cos 11^{\circ}+\sin 11^{\circ}} is equal to:
View answer and explanation
Correct option: A
cot⁡56∘\cot 56^{\circ}
Option A (cot⁡56∘\cot 56^{\circ}) is correct because the expression simplifies to cot⁡56∘\cot 56^{\circ} by dividing by cos⁡11∘\cos 11^{\circ} and using the tangent subtraction formula.
131 Easy
131 of 394
Application +4 | -1
If cos⁡A3=cos⁡B4=15,−π/2<A,B<0\frac{\cos A}{3}=\frac{\cos B}{4}=\frac{1}{5},-\pi / 2< A, B < 0, then value of 2sin⁡A+4sin⁡B2 \sin A+4 \sin B is equal to
View answer and explanation
Correct option: C
-4
Option C (-4) is correct because by finding sinA and sinB using the given information and substituting them into the expression, the result is -4.
132 Easy
132 of 394
Application +4 | -1
If tan⁡θ=1/3\tan \theta=1 / \sqrt{3}, then the value of 3cos⁡2θ+sin⁡2θ\sqrt{3} \cos 2 \theta+\sin 2 \theta is
View answer and explanation
Correct option: A
3\sqrt{3}
Option A (3\sqrt{3}) is correct because substituting tan⁡θ=1/3\tan \theta = 1/\sqrt{3} into the expression 3cos⁡2θ+sin⁡2θ\sqrt{3} \cos 2 \theta+\sin 2 \theta yields 3\sqrt{3}.
133 Medium
133 of 394
Application +4 | -1
If ABCA B C is an equilateral triangle such that sin⁡A+sin⁡B+sin⁡C=kcos⁡A2cos⁡B2cos⁡C2\sqrt{\sin A}+\sqrt{\sin B}+\sqrt{\sin C}=\sqrt{k \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}}. Then find the value of &#39; kk &#39;.
View answer and explanation
Correct option: A
12
Option A (12) is correct because substituting A=B=C=60 degrees into the given equation and simplifying leads to k=12.
134 Easy
134 of 394
Application +4 | -1
In tan⁡  θ+sec⁡  θ=3,   0<θ<π\tan \,\,\theta + \sec \,\,\theta = \sqrt 3 ,\;\,0 < \theta < \pi, then θ\theta is equal to

View answer and explanation
Correct option: C
π/6\pi /6
Option C (π/6\pi /6) is correct because solving the given equation tan⁡θ+sec⁡θ=3\tan \theta + \sec \theta = \sqrt 3 leads to this value within the specified range.
135 Easy
135 of 394
Application +4 | -1
If A=130∘A=130^{\circ} and x=sin⁡A+cos⁡Ax=\sin A+\cos A, then
View answer and explanation
Correct option: A
x>0x > 0
The correct answer is A (x>0x > 0) because when A=130∘A = 130^{\circ}, x=sin⁡A+cos⁡A=sin⁡130∘+cos⁡130∘=sin⁡50∘−cos⁡50∘x = \sin A + \cos A = \sin 130^{\circ} + \cos 130^{\circ} = \sin 50^{\circ} - \cos 50^{\circ}, and since sin⁡50∘>∣cos⁡50∘∣\sin 50^{\circ} > |\cos 50^{\circ}|, xx is positive.

Explore Other Chapters in MATHEMATICS

View All →
SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs
136 Easy
136 of 394
Application +4 | -1
If acos⁡A−bsin⁡A=ca \cos A-b \sin A=c, then asin⁡A+bcos⁡Aa \sin A+b \cos A is equal to
View answer and explanation
Correct option: A
±a2+b2−c2\pm \sqrt{a^2+b^2-c^2}
Option A (±a2+b2−c2\pm \sqrt{{a^2+b^2-c^2}}) is correct because it is derived by squaring both the given equation and the expression to be found, then adding them.
137 Easy
137 of 394
Application +4 | -1
The equation 8sec⁡2θ−6sec⁡θ+1=08 \sec ^2 \theta-6 \sec \theta+1=0 has -
View answer and explanation
Correct option: D
No roots
Option D (No roots) is correct because the equation leads to sec⁡θ\sec \theta values that are not possible, as the absolute value of sec⁡θ\sec \theta must be greater than or equal to 1.
138 Easy
138 of 394
Application +4 | -1
If f(x)=cos⁡2  x+sec⁡2  xf\left( x \right) = {\cos ^2}\,\,x + {\sec ^2}\,\,x, its value always is
View answer and explanation
Correct option: B
f(x)=1f\left( x \right) = 1
Option D (f(x)≥2f(x) \ge 2) is correct because the minimum value of cos⁡2x+sec⁡2x{\cos ^2}x + {\sec ^2}x occurs when cos⁡2x=1\cos^2 x = 1 and sec⁡2x=1\sec^2 x = 1, so the minimum value is 2.
139 Easy
139 of 394
Analyzing +4 | -1
Let nn be a positive integer such that sin⁡  π2n+cos⁡  π2n=n2\sin \,\,\frac{\pi }{{2n}} + \cos \,\,\frac{\pi }{{2n}} = \frac{{\sqrt n }}{2}. Then
View answer and explanation
Correct option: A
n=6n = 6
Option A (n=6n = 6) is correct because it is the only integer solution that satisfies the given trigonometric equation after squaring and simplifying.
140 Medium
140 of 394
Application +4 | -1
If in a △ABC,2cos⁡Asin⁡C=sin⁡B\triangle A B C, 2 \cos A \sin C=\sin B, then the triangle is
View answer and explanation
Correct option: B
isosceles
Option B (isosceles) is correct because the given condition leads to the conclusion that two angles of the triangle are equal, making it isosceles.
141 Medium
141 of 394
Application +4 | -1
If 1+sin⁡x+sin⁡2x+sin⁡3x+…∞1+\sin x+\sin ^2 x+\sin ^3 x+\ldots \infty is equal to 4+23,0<x<π4+2 \sqrt{3}, 0 < x < \pi then xx is equal to -
View answer and explanation
Correct option: D
π3\frac{\pi}{3} or 2π3\frac{2 \pi}{3}
Option A (a=2c\mathrm{a}=\sqrt{{2}} \mathrm{c}) is correct because equating the perpendicular distance from the center of the circle to the tangent and the y-intercept of the tangent to the parabola gives this relationship.
142 Easy
142 of 394
Application +4 | -1
The maximum value of cos⁡2(π3−x)  −cos⁡2(π3+x){\cos ^2}\left( {\frac{\pi }{3} - x} \right)\,\, - {\cos ^2}\left( {\frac{\pi }{3} + x} \right) is
View answer and explanation
Correct option: C
32\frac{{\sqrt 3 }}{2}
Option C is correct because the maximum value is found by simplifying the expression and using the maximum value of the sine function.
143 Easy
143 of 394
Application +4 | -1
log⁡tan⁡1∘+log⁡tan⁡2∘+…..+log⁡tan⁡89∘\log \tan 1^{\circ}+\log \tan 2^{\circ}+\ldots . .+\log \tan 89^{\circ} is
View answer and explanation
Correct option: B
0
Option B (0) is correct because log⁡tan⁡1∘+log⁡tan⁡2∘+⋯+log⁡tan⁡89∘=log⁡(tan⁡1∘⋅tan⁡2∘⋅⋯⋅tan⁡89∘)\log \tan 1^{\circ} + \log \tan 2^{\circ} + \dots + \log \tan 89^{\circ} = \log (\tan 1^{\circ} \cdot \tan 2^{\circ} \cdot \dots \cdot \tan 89^{\circ}). Since tan⁡θ⋅tan⁡(90∘−θ)=1\tan \theta \cdot \tan (90^{\circ} - \theta) = 1, the product inside the logarithm is 1, and log⁡1=0\log 1 = 0.
144 Easy
144 of 394
Application +4 | -1
If tan⁡A=12\tan A=\frac{1}{2}, then tan⁡3A=\tan 3 A=
View answer and explanation
Correct option: B
112\frac{11}{2}
Option B (112\frac{{11}}{{2}}) is correct because using the triple angle formula for tangent, tan⁡3A\tan 3A is calculated as 112\frac{{11}}{{2}}.
145 Medium
145 of 394
Application +4 | -1
If (sec⁡α+tan⁡α)(sec⁡β+tan⁡β)(sec⁡γ+tan⁡γ)(\sec \alpha+\tan \alpha)(\sec \beta+\tan \beta)(\sec \gamma+\tan \gamma) =tan⁡αtan⁡βtan⁡γ=\tan \alpha \tan \beta \tan \gamma, then expression (sec⁡α−tan⁡α)(sec⁡β−tan⁡β)(sec⁡γ−tan⁡γ)(\sec \alpha-\tan \alpha)(\sec \beta-\tan \beta)(\sec \gamma-\tan \gamma) is equal to
View answer and explanation
Correct option: A
cot⁡αcot⁡βcot⁡γ\cot \alpha \cot \beta \cot \gamma
Option A (cot⁡αcot⁡βcot⁡γ\cot \alpha \cot \beta \cot \gamma) is correct because using the identity sec2x−tan2x=1sec^2 x - tan^2 x = 1 and the given equation, we can derive the expression.
146 Easy
146 of 394
Application +4 | -1
If sin⁡x  cos⁡x  cos⁡  2x=λ\sin x\,\,\cos x\,\,\cos \,\,2x = \lambda has a solution, then λ\lambda lies in the interval
View answer and explanation
Correct option: A
[−1/4,   1/4]\left[ { - 1/4,\;\,1/4} \right]
Option A is correct because the range of λ\lambda is found by considering the range of the sine function.
147 Hard
147 of 394
Analyzing +4 | -1
If ∣ cos⁡  θ  { sin⁡  θ+sin⁡2  θ+sin⁡2  α} ∣≤k\left| {\,\cos \,\,\theta \,\,\left\{ {\,\sin \,\,\theta + \sqrt {{{\sin }^2}\,\,\theta + {{\sin }^2}\,\,\alpha } } \right\}\,} \right| \le k, then the value of kk is
View answer and explanation
Correct option: B
1+sin⁡2  α\sqrt {1 + {{\sin }^2}\,\,\alpha }
Option B (1+sin⁡2 α\sqrt {1 + {{\sin }^2}\,\alpha }) is correct because it represents the maximum value of the given expression, derived using trigonometric manipulations and inequalities.
148 Medium
148 of 394
Application +4 | -1
The value of cos⁡(2π7)+cos⁡(4π7)+cos⁡(6π7)\cos \left( {\frac{{2\pi }}{7}} \right) + \cos \left( {\frac{{4\pi }}{7}} \right) + \cos \left( {\frac{{6\pi }}{7}} \right) is equal to:
View answer and explanation
Correct option: B
−12- \frac{1}{2}
Option B (-1/2) is correct because the sum of cosines can be simplified using trigonometric identities to obtain the result -1/2.
149 Medium
149 of 394
Application +4 | -1
Find the value of kk for which the equation 4cos⁡36∘+cot⁡(152)∘=∑r=1kr4 \cos 36^{\circ}+\cot \left(\frac{15}{2}\right)^{\circ}=\sum\limits_{{r=1}}^k \sqrt{r} is satisfied.
View answer and explanation
Correct option: A
6
Option A (6) is correct because by substituting the known values of cos⁡36∘\cos 36^{\circ} and cot⁡(152)∘\cot (\frac{15}{2})^{\circ}, the equation simplifies to ∑r=16r\sum_{r=1}^6 \sqrt{r}, thus k=6.
150 Easy
150 of 394
Application +4 | -1
If sin⁡  x+sin⁡2  x=1\sin \,\,x + {\sin ^2}\,\,x = 1, then cos⁡12  x+3  cos⁡10  x+3  cos⁡8  x+cos⁡6  x{\cos ^{12}}\,\,x + 3\,\,{\cos ^{10}}\,\,x + 3\,\,{\cos ^8}\,\,x + {\cos ^6}\,\,x is equal to
View answer and explanation
Correct option: A
1\mathrm{1}
Option B (tan⁡ B−tan⁡ C−tan⁡ A\tan \,B - \tan \,C - \tan \,A) is correct because it is derived by applying the tangent addition formula and rearranging the terms.
Revision notes and key takeaways

Review TRIGONOMETRY RATIOS in your textbook or class notes, then use this set to check your understanding.

Showing 121-150 of 394 questions

Continue Your Practice

📚 More Sets in This Chapter

Try different difficulty levels to master this topic

→ ENGINEERING MATHEMATICS - TRIGONOMETRY RATIOS MCQs (Easy) 218 MCQs → ENGINEERING MATHEMATICS - TRIGONOMETRY RATIOS MCQs (Hard) 19 MCQs → ENGINEERING MATHEMATICS - TRIGONOMETRY RATIOS MCQs (Medium) 157 MCQs
🎯 Next Chapters to Practice

Continue building your subject mastery

→ SEQUENCE OF SERIES 1764 MCQs → DEFINITE INTEGRATION 1443 MCQs → MATRICES 1363 MCQs → INDEFINITE INTEGRATION 1179 MCQs

Your practice

0 / 394 answered · 394 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Practice settings

Type at least 2 characters or jump directly:
🩺 NEET MCQs 🔬 JEE MCQs 🎯 JEE College Predictor Study guides 📊 Dashboard
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.

Continue with Free Account →