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JEE • MATHEMATICS • TRANSFORMATIONS

Transformations practice · Medium

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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 31-45 of 45 questions
31 Medium
31 of 45
Application +4 | -1
Given π2<α<π\frac{\pi}{2} < \alpha < \pi, then the expression 1−sin⁡α1+sin⁡α+1+sin⁡α1−sin⁡α\sqrt{\frac{1-\sin \alpha}{1+\sin \alpha}}+\sqrt{\frac{1+\sin \alpha}{1-\sin \alpha}} is equal to
View answer and explanation
Correct option: B
−2cos⁡α-\frac{2}{\cos \alpha}
Option B (−2cos⁡α-\frac{2}{\cos \alpha}) is correct because, given the range of α\alpha, the expression simplifies to this value using trigonometric identities.
32 Medium
32 of 45
Application +4 | -1
If A+B+C=270∘A + B + C = 270^\circ, then

cos⁡  2A+cos⁡  2B+cos⁡  2C\cos \,\,2A + \cos \,\,2B + \cos \,\,2C is equal to
View answer and explanation
Correct option: C
1−4  sin⁡A  sin⁡B  sin⁡C  1 - 4\,\,\sin A\,\,\sin B\,\,\sin C\;
Option D (1−4 cos⁡A cos⁡B cos⁡C1 - 4\,\cos A\,\cos B\,\cos C) is correct because it is the result of applying trigonometric identities to simplify the given expression with the condition A+B+C=270∘A + B + C = 270^\circ.
33 Medium
33 of 45
Analyzing +4 | -1
If cos⁡  (α+β)sin⁡  (γ+δ)=cos⁡  (α−β)sin⁡  (γ−δ)\cos \,\,\left( {\alpha + \beta } \right)\sin \,\,\left( {\gamma + \delta } \right) = \cos \,\,\left( {\alpha - \beta } \right)\sin \,\,\left( {\gamma - \delta } \right), then cot⁡α  cot⁡β  cot⁡γ\cot \alpha \,\,\cot \beta \,\,\cot \gamma is equal to
View answer and explanation
Correct option: A
cot⁡  δ\cot \,\,\delta
Option A (cot⁡δ\cot \delta) is correct because after applying trigonometric identities and simplifying the given equation, we arrive at the relation cot⁡αcot⁡βcot⁡γ=cot⁡δ\cot \alpha \cot \beta \cot \gamma = \cot \delta.
34 Medium
34 of 45
Application +4 | -1
If cos⁡  (θ+ϕ)=m  cos⁡  (θ−ϕ)\cos \,\,\left( {\theta + \phi } \right) = m\,\,\cos \,\,\left( {\theta - \phi } \right) is equal to
View answer and explanation
Correct option: C
1−m1+m  cot⁡  ϕ\frac{{1 - m}}{{1 + m}}\,\,\cot \,\,\phi
Option C (1−m1+mcot⁡ϕ\frac{{1 - m}}{{1 + m}}\cot \phi) is correct because it is the result of applying the componendo and dividendo rule and trigonometric identities to the given equation.
35 Medium
35 of 45
Application +4 | -1
If − π≤x≤π,   − π≤y≤π   and  cos⁡  x+cos⁡  y=2- \,\pi \le x \le \pi ,\;\, - \,\pi \le y \le \pi \;\,\mathrm{and}\,\,\cos \,\,x + \cos \,\,y = 2, then the value of cos(x−y)\mathrm{cos}\left( {x - y} \right) is
View answer and explanation
Correct option: C
1
Option C (1) is correct because the maximum value of cosine is 1, so both cos(x) and cos(y) must be 1. This implies x = 0 and y = 0, therefore cos(x - y) = cos(0) = 1.
36 Medium
36 of 45
Application +4 | -1
The value of 1+cos⁡  56∘+cos⁡  58∘−cos⁡  66∘1 + \cos \,\,56^\circ + \cos \,\,58^\circ - \cos \,\,66^\circ is
View answer and explanation
Correct option: A
4  cos⁡  28∘  cos⁡  29∘  sin⁡33∘4\,\,\cos \,\,28^\circ \,\,\cos \,\,29^\circ \,\,\sin 33^\circ
Option A (4cos⁡28∘cos⁡29∘sin⁡33∘4\cos 28^\circ \cos 29^\circ \sin 33^\circ) is correct because it represents the simplified form of the trigonometric expression after applying the sum-to-product and other trigonometric identities.
37 Medium
37 of 45
Analyzing +4 | -1
Which of the following functions have period 2π2\mathrm{\pi }?
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Correct option: C
y=sin⁡t+cos⁡2ty = \sin t + \cos 2t
Option C is correct. The period of sin⁡t\sin t is 2π2\pi and the period of cos⁡2t\cos 2t is π\pi. The least common multiple of 2π2\pi and π\pi is 2π2\pi.
38 Medium
38 of 45
Application +4 | -1
The value of sin⁡  (B+A)+  cos  (B−A)sin⁡  (B−A)+cos⁡  (B+A)\frac{{\sin \,\,\left( {B + A} \right) + \;\mathrm{cos}\,\,\left( {B - A} \right)}}{{\sin \,\,\left( {B - A} \right) + \cos \,\,\left( {B + A} \right)}} is equal to
View answer and explanation
Correct option: B
cos⁡  A+sin⁡  Acos⁡  A−sin⁡  A\frac{{\cos \,\,A + \sin \,\,A}}{{\cos \,\,A - \sin \,\,A}}
Option B (cos⁡,A+sin⁡,Acos⁡,A−sin⁡,A\frac{{\cos \\,A + \sin \\,A}}{{\cos \\,A - \sin \\,A}}) is correct because simplifying the given expression using trigonometric identities leads to the result cos⁡,A+sin⁡,Acos⁡,A−sin⁡,A\frac{{\cos \\,A + \sin \\,A}}{{\cos \\,A - \sin \\,A}}.
39 Medium
39 of 45
Analyzing +4 | -1
In a △ABC,a,b\triangle A B C, a, b and cc are the sides of the triangle opposite to the angles 2222222222 dA,B,C2222222222 \mathrm{~d} A, B, C, respectively. Then, the value of a3sin⁡(B−C)+b3sin⁡(C−A)+c3sin⁡(A−B)a^3 \sin (B-C)+b^3 \sin (C-A)+c^3 \sin (A-B) is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the expression simplifies to zero using trigonometric identities and the sine rule.
40 Medium
40 of 45
Application +4 | -1
If f:R→Rf:R \to R and is defined by f(x)=12−cos⁡3xf\left( x \right) = \frac{1}{{2 - \cos 3x}} for each x∈Rx \in R, then the range ff is
View answer and explanation
Correct option: B
[1/3,1]\left[ {1/3,\mathrm{ }1} \right]
Option B [1/3,1]\left[ {1/3,\mathrm{ }1} \right] is correct because the range is found by considering the bounds of the cosine function and its effect on the given function.
41 Medium
41 of 45
Application +4 | -1
If A+B+C  =3 π2A + B + C\; = \frac{{3\,\pi }}{2}, then cos⁡  2A+cos⁡  2B+cos⁡  2C\cos \,\,2A + \cos \,\,2B + \cos \,\,2C is equal to
View answer and explanation
Correct option: D
1−4sin⁡A  sin⁡B  sin⁡C1 - 4\sin A\,\,\sin B\,\,\sin C
Option D (1−4sin⁡Asin⁡Bsin⁡C1 - 4\sin A\sin B\sin C) is correct because, with the given condition, the expression simplifies to this form using trigonometric identities.
42 Medium
42 of 45
Analyzing +4 | -1
If y=cos⁡2x+sec⁡2xy = {\cos ^2}x + {\sec ^2}x, then
View answer and explanation
Correct option: C
y≥2y \ge 2
Option C (y≥2y \ge 2) is correct because cos⁡2x+sec⁡2x{\cos ^2}x + {\sec ^2}x can be rewritten as (cos⁡x−sec⁡x)2+2(\cos x - \sec x)^2 + 2, which is always greater than or equal to 2.
43 Medium
43 of 45
Analyzing +4 | -1
If   A+B+C=π    and  cos⁡  A=cos⁡  B  cos⁡  C\;A + B + C = \pi \,\,\;\mathrm{and}\,\,\cos \,\,A = \cos \,\,B\,\,\cos \,\,C then tan⁡  B  tan⁡  C\tan \,\,B\,\,\tan \,\,C is equal to

View answer and explanation
Correct option: B
1\mathrm{1}
Option B (1) is correct because by using the given condition A+B+C=πA + B + C = \pi and the trigonometric identity, we can derive that tan⁡Btan⁡C=1\tan B \tan C = 1.
44 Medium
44 of 45
Analyzing +4 | -1
If the equation sin⁡  θ(sin⁡  θ+2  cos⁡  θ)=a\sin \,\,\theta \left( {\sin \,\,\theta + 2\,\,\cos \,\,\theta } \right) = a has a real solution, then the shortest interval containing ′a′'a' is
View answer and explanation
Correct option: A
[1−52,  1+52]\left[ {\frac{{1 - \sqrt 5 }}{2},\,\,\frac{{1 + \sqrt 5 }}{2}} \right]
Option A is correct because the range of 'a' is determined by finding the maximum and minimum values of the trigonometric expression.
45 Medium
45 of 45
Application +4 | -1
If A+B+C=πA + B + C = \pi, then sin⁡  2A+sin⁡  2B+sin⁡  2C\sin \,\,2A + \sin \,\,2B + \sin \,\,2C is equal to
View answer and explanation
Correct option: A
4  sin⁡A  sin⁡B  sin⁡C4\,\,\sin A\,\,\sin B\,\,\sin C
Option A (4  sin⁡A  sin⁡B  sin⁡C4\,\,\sin A\,\,\sin B\,\,\sin C) is correct because it is the result of applying trigonometric identities to simplify the expression sin⁡2A+sin⁡2B+sin⁡2C\sin 2A + \sin 2B + \sin 2C when A+B+C=πA + B + C = \pi.
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