Trigonometry · JEE Main previous year questions

2025 18 questions

1. Q92025 · 8 Apr 2025 · Shift 2

   The value of ${\cot }^{-1}\left(\frac{\sqrt{1+{\tan }^2(2)}-1}{\tan (2)}\right)-{\cot }^{-1}\left(\frac{\sqrt{1+{\tan }^2\left(\frac{1}{2}\right)}+1}{\tan \left(\frac{1}{2}\right)}\right)$ is equal to

   1. A.$\pi -\frac{3}{2}$
   2. B.$\pi +\frac{5}{2}$
   3. C.$\pi -\frac{5}{4}$
   4. D.$\pi +\frac{3}{2}$
   Show answer

   Answer: C

2. Q102025 · 8 Apr 2025 · Shift 2

   Let $A=\left\{\theta \in [0,2\pi ]:1+10\mathrm{Re}\left(\frac{2\cos \theta +i\sin \theta }{\cos \theta -3i\sin \theta }\right)=0\right\}$. Then ${\sum }_{\theta \in A}{\theta }^2$ is equal to

   1. A.$\frac{21}{4}{\pi }^2$
   2. B.$6{\pi }^2$
   3. C.$\frac{27}{4}{\pi }^2$
   4. D.$8{\pi }^2$
   Show answer

   Answer: A

3. Q82025 · 4 Apr 2025 · Shift 1

   If $10{\sin }^4\theta +15{\cos }^4\theta =6$, then the value of $\frac{27{\mathrm{cosec}}^6\theta +8{\sec }^6\theta }{16{\sec }^8\theta }$ is

   1. A.$\frac{2}{5}$
   2. B.$\frac{3}{5}$
   3. C.$\frac{1}{5}$
   4. D.$\frac{3}{4}$
   Show answer

   Answer: A

4. Q202025 · 4 Apr 2025 · Shift 1

   Considering the principal values of the inverse trigonometric functions, ${\sin }^{-1}\left(\frac{\sqrt{3}}{2}x+\frac{1}{2}\sqrt{1-x^2}\right),-\frac{1}{2}<x<\frac{1}{\sqrt{2}}$, is equal to

   1. A.$-\frac{5\pi }{6}-{\sin }^{-1}x$
   2. B.$\frac{5\pi }{6}-{\sin }^{-1}x$
   3. C.$\frac{\pi }{6}+{\sin }^{-1}x$
   4. D.$\frac{\pi }{4}+{\sin }^{-1}x$
   Show answer

   Answer: C

5. Q142025 · 4 Apr 2025 · Shift 2

   Let A be the point of intersection of the lines ${\mathrm{L}}_1:\frac{x-7}{1}=\frac{y-5}{0}=\frac{z-3}{-1}$ and ${\mathrm{L}}_2:\frac{x-1}{3}=\frac{y+3}{4}=\frac{z+7}{5}$. Let B and C be the points on the lines ${\mathrm{L}}_1$ and ${\mathrm{L}}_2$ respectively such that $AB=AC=\sqrt{15}$. Then the square of the area of the triangle $ABC$ is :

   1. A.63
   2. B.57
   3. C.60
   4. D.54
   Show answer

   Answer: D

6. Q42025 · 3 Apr 2025 · Shift 1

   The number of solutions of the equation $2x+3\tan x=\pi$, $x\in [-2\pi ,2\pi ]-\left\{\pm \frac{\pi }{2},\pm \frac{3\pi }{2}\right\}$ is:

   1. A.4
   2. B.5
   3. C.3
   4. D.6
   Show answer

   Answer: B

7. Q182025 · 3 Apr 2025 · Shift 1

   If $y(x)=\left|\begin{array}{ccc}\sin x& \cos x& \sin x+\cos x+1\\ 27& 28& 27\\ 1& 1& 1\end{array}\right|$, $x\in \mathbb{R}$, then $\frac{d^{2}y}{d{x}^2}+y$ is equal to

   1. A.28
   2. B.27
   3. C.-1
   4. D.1
   Show answer

   Answer: C

8. Q162025 · 3 Apr 2025 · Shift 2

   The number of solutions of the equation $(4-\sqrt{3})\sin x-2\sqrt{3}{\cos }^{2}x=-\frac{4}{1+\sqrt{3}},x\in \left[-2\pi ,\frac{5\pi }{2}\right]$ is

   1. A.$4$
   2. B.$3$
   3. C.$6$
   4. D.$5$
   Show answer

   Answer: D

9. Q42025 · 2 Apr 2025 · Shift 1

   If $\theta \in [-2\pi ,2\pi ]$, then the number of solutions of $2\sqrt{2}{\cos }^2\theta +(2-\sqrt{6})\cos \theta -\sqrt{3}=0$, is equal to:

   1. A.8
   2. B.6
   3. C.10
   4. D.12
   Show answer

   Answer: A

10. Q92025 · 2 Apr 2025 · Shift 2

    If $\theta \in [-\frac{7\pi }{6},\frac{4\pi }{3}]$, then the number of solutions of $\sqrt{3}{\mathrm{cosec}}^2\theta -2(\sqrt{3}-1)\mathrm{cosec}\theta -4=0$, is equal to :

    1. A.7
    2. B.10
    3. C.6
    4. D.8
    Show answer

    Answer: C

11. Q212025 · 2 Apr 2025 · Shift 2· Numerical answer

    If $y=\cos \left(\frac{\pi }{3}+{\cos }^{-1}\frac{x}{2}\right)$, then $(x-y{)}^2+3y^2$ is equal to ____.

    Show answer

    Answer: 3

12. Q152025 · 29 Jan 2025 · Shift 2

    If $\sin x+{\sin }^{2}x=1$, $x\in \left(0,\frac{\pi }{2}\right)$, then $({\cos }^{12}x+{\tan }^{12}x)+3({\cos }^{10}x+{\tan }^{10}x+{\cos }^{8}x+{\tan }^{8}x)+({\cos }^{6}x+{\tan }^{6}x)$ is equal to:

    1. A.$3$
    2. B.$4$
    3. C.$2$
    4. D.$1$
    Show answer

    Answer: C

13. Q62025 · 28 Jan 2025 · Shift 1

    $\cos \left({\sin }^{-1}\frac{3}{5}+{\sin }^{-1}\frac{5}{13}+{\sin }^{-1}\frac{33}{65}\right)$ is equal to:

    1. A.$\frac{33}{65}$
    2. B.1
    3. C.$\frac{32}{65}$
    4. D.0
    Show answer

    Answer: D

14. Q92025 · 28 Jan 2025 · Shift 1

    Let ${}^n{C}_{r-1}=28,{}^n{C}_r=56$ and ${}^n{C}_{r+1}=70$. Let $A(4\cos t,4\sin t),B(2\sin t,-2\cos t)$ and $C\left(3r-n,r^2-n-1\right)$ be the vertices of a triangle $ABC$, where $t$ is a parameter. If $(3x-1{)}^2+(3y{)}^2=\alpha$, is the locus of the centroid of triangle ABC , then $\alpha$ equals

    1. A.18
    2. B.8
    3. C.20
    4. D.6
    Show answer

    Answer: C

15. Q242025 · 24 Jan 2025 · Shift 1· Numerical answer

    If for some $\alpha ,\beta ;\alpha \le \beta ,\alpha +\beta =8$ and ${\sec }^2\left({\tan }^{-1}\alpha \right)+{\mathrm{cosec}}^2\left({\cot }^{-1}\beta \right)=36$, then ${\alpha }^2+\beta$ is ____

    Show answer

    Answer: 14

16. Q42025 · 24 Jan 2025 · Shift 2

    Let $\mathrm{A}=\left\{x\in (0,\pi )-\left\{\frac{\pi }{2}\right\}:{\log }_{(2/\pi )}|\sin x|+{\log }_{(2/\pi )}|\cos x|=2\right\}$ and $\mathrm{B}=\{x\ge 0:\sqrt{x}(\sqrt{x}-4)-3|\sqrt{x}-2|+6=0\}$. Then $\mathrm{n}(\mathrm{A}\cup \mathrm{B})$ is equal to :

    1. A.4
    2. B.8
    3. C.6
    4. D.2
    Show answer

    Answer: B

17. Q72025 · 24 Jan 2025 · Shift 2

    If $\alpha >\beta >\gamma >0$, then the expression ${\cot }^{-1}\left\{\beta +\frac{(1+{\beta }^2)}{(\alpha -\beta )}\right\}+{\cot }^{-1}\left\{\gamma +\frac{(1+{\gamma }^2)}{(\beta -\gamma )}\right\}+{\cot }^{-1}\left\{\alpha +\frac{(1+{\alpha }^2)}{(\gamma -\alpha )}\right\}$ is equal to :

    1. A.$3\pi$
    2. B.$\frac{\pi }{2}-(\alpha +\beta +\gamma )$
    3. C.$\pi$
    4. D.0
    Show answer

    Answer: C

18. Q182025 · 22 Jan 2025 · Shift 1

    Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum values of $16\left(({\sec }^{-1}x{)}^2+({\mathrm{cosec}}^{-1}x{)}^2\right)$ is :

    1. A.$24{\pi }^2$
    2. B.$18{\pi }^2$
    3. C.$22{\pi }^2$
    4. D.$31{\pi }^2$
    Show answer

    Answer: C