Q.Find the value of the following:

Showing the 12 most recent of 14 on this concept.

1. GUJCET 2019Set 171 markMCQ
   Q.$\begin{vmatrix} \sin^2\theta & \cos^2\theta \\ -\cos^2\theta & \sin^2\theta \end{vmatrix} = \underline{\hspace{2cm}}$. (A) $\dfrac{1}{2}(1 + \cos^2 2\theta)$ (B) $\dfrac{1}{2}(1 - \sin^2 2\theta)$ (C) $\cos 2\theta$ (D) $\dfrac{1}{2}\sin^2 2\theta$

   ›Reveal solutionSolution

   $\begin{vmatrix}\sin^2\theta&\cos^2\theta\\-\cos^2\theta&\sin^2\theta\end{vmatrix}=\sin^4\theta+\cos^4\theta$, which equals $\tfrac12(1+\cos^2 2\theta)$.

   Concept: Expand: $\sin^2\theta\cdot\sin^2\theta-\cos^2\theta\cdot(-\cos^2\theta)=\sin^4\theta+\cos^4\theta$.

   Now $\sin^4\theta+\cos^4\theta=1-2\sin^2\theta\cos^2\theta=1-\tfrac12\sin^2 2\theta$. Using $\sin^2 2\theta=1-\cos^2 2\theta$:

   $1-\tfrac12(1-\cos^2 2\theta)=\tfrac12+\tfrac12\cos^2 2\theta=\tfrac12(1+\cos^2 2\theta)$

   (Check $\theta=0$: determinant $=1$, and $\tfrac12(1+1)=1$.)

   ✓Final answer

   (A) $\dfrac{1}{2}(1+\cos^2 2\theta)$

   ANSWER: (A)

2. GUJCET 2025Set 031 markMCQ
   Q.$\begin{vmatrix} \cos^2\theta & -\sin^2\theta \\ \sin^2\theta & \cos^2\theta \end{vmatrix} =$ _____. (A) $\frac{1}{2} - \frac{1}{2}\cos^2 2\theta$ (B) $\frac{1}{4}(3 + \cos 4\theta)$ (C) $1 + \frac{1}{2}\sin^2 2\theta$ (D) $1 + 2\sin^2\theta \cdot \cos^2\theta$

   ›Reveal solutionSolution

   Expand the 2×2 determinant, then use double/quadruple-angle identities.

   $\begin{vmatrix}\cos^2\theta&-\sin^2\theta\\\sin^2\theta&\cos^2\theta\end{vmatrix}=\cos^4\theta+\sin^4\theta=1-2\sin^2\theta\cos^2\theta=1-\tfrac12\sin^2 2\theta.$

   Using $\sin^2 2\theta=\tfrac{1-\cos4\theta}{2}$:

   $1-\tfrac12\cdot\tfrac{1-\cos4\theta}{2}=\frac{4-1+\cos4\theta}{4}=\frac{3+\cos4\theta}{4}.$

   ✓Final answer

   (B) $\dfrac14(3+\cos4\theta)$

   ANSWER: (B)

3. GUJCET 2020Set 071 markMCQ
   Q.For $\triangle ABC$, the value of $\begin{vmatrix} 0 & \sin A & \tan B \\ -\sin(B + C) & 0 & \cos C \\ \tan(A + C) & -\cos C & 0 \end{vmatrix} =$ ________. (A) $-1$ (B) $0$ (C) $1$ (D) $\sin A \cos C$

   ›Reveal solutionSolution

   Using $A+B+C=\pi$ the matrix becomes skew-symmetric; a $3\times3$ (odd-order) skew-symmetric determinant is always $0$.

   Concept — trig identities in a triangle. Since $A+B+C=\pi$: $\sin(B+C)=\sin(\pi-A)=\sin A$ and $\tan(A+C)=\tan(\pi-B)=-\tan B$.

   Substituting, the matrix is

   $\begin{vmatrix} 0 & \sin A & \tan B \\ -\sin A & 0 & \cos C \\ -\tan B & -\cos C & 0 \end{vmatrix}$

   Every $a_{ij}=-a_{ji}$ with zero diagonal, i.e. it is skew-symmetric. For any odd-order skew-symmetric matrix $\det = 0$.

   ✓Final answer

   (B) $0$

   ANSWER: (B)

4. GUJCET 2023Set 091 markMCQ
   Q.$\begin{vmatrix} \sin\frac{11\pi}{36} & \cos\frac{11\pi}{36} \\ \sin\frac{2\pi}{9} & \cos\frac{2\pi}{9} \end{vmatrix} =$ ______. (A) $\cos\frac{\pi}{12}$ (B) $\sin\frac{2\pi}{9}$ (C) $\cos\frac{5\pi}{12}$ (D) $\sin\frac{7\pi}{12}$

   ›Reveal solutionSolution

   A determinant of this sin/cos form collapses to a single sine of the angle difference.

   Concept. $\begin{vmatrix}\sin P&\cos P\\\sin Q&\cos Q\end{vmatrix}=\sin P\cos Q-\cos P\sin Q=\sin(P-Q)$.

   Solution. $P=\tfrac{11\pi}{36}$, $Q=\tfrac{2\pi}{9}=\tfrac{8\pi}{36}$.

   $\sin\!\left(\tfrac{11\pi}{36}-\tfrac{8\pi}{36}\right)=\sin\tfrac{3\pi}{36}=\sin\tfrac{\pi}{12}.$

   Since $\cos\tfrac{5\pi}{12}=\cos 75^\circ=\sin 15^\circ=\sin\tfrac{\pi}{12}$, the value equals $\cos\tfrac{5\pi}{12}$.

   ✓Final answer

   (C) $\cos\frac{5\pi}{12}$

   ANSWER: (C)

5. GUJCET 2020Set 071 markMCQ
   Q.Let $f(t) = \begin{vmatrix} \cos t & t & 1 \\ 2\tan t & t & 2t \\ \tan t & t & t \end{vmatrix}$. Then $\lim_{t \to 0} \frac{f(t)}{t^2}$ is equal to ________. (A) $3$ (B) $1$ (C) $-1$ (D) $0$

   ›Reveal solutionSolution

   The determinant equals $t(-t\cos t+\tan t)$, so $f(t)/t^2 = -\cos t + \tfrac{\tan t}{t}\to -1+1 = 0$.

   Concept — simplify the determinant first. Column 2 is $t[1,1,1]^T$, so pull out $t$:

   $f(t) = t\,\begin{vmatrix}\cos t & 1 & 1\\ 2\tan t & 1 & 2t\\ \tan t & 1 & t\end{vmatrix} = t\, g(t)$

   Expand $g(t)$ along the first column's cofactors (about row 1):

   $g(t) = \cos t\,(t-2t) - 1\,(2t\tan t - 2t\tan t) + 1\,(2\tan t-\tan t)$

   $= -t\cos t + 0 + \tan t$

   Therefore

   $\frac{f(t)}{t^2} = \frac{t\,g(t)}{t^2} = \frac{g(t)}{t} = -\cos t + \frac{\tan t}{t}$

   Taking $t\to 0$ (using $\tfrac{\tan t}{t}\to 1$):

   $\lim_{t\to 0}\frac{f(t)}{t^2} = -1 + 1 = 0$

   ✓Final answer

   Option (D) $0$

   ANSWER: (D)

6. GUJCET 2022Set 081 markMCQ
   Q.For real numbers $x,y,z$ such that $x\ne y\ne z$, $\begin{vmatrix}x & x^2 & 1+x^3\\ y & y^2 & 1+y^3\\ z & z^2 & 1+z^3\end{vmatrix}=0$ and $\begin{vmatrix}1 & x & x^2\\ 1 & y & y^2\\ 1 & z & z^2\end{vmatrix}\ne 0$ then $xyz=$ ______. (A) $2$ (B) $-1$ (C) $0$ (D) $1$

   ›Reveal solutionSolution

   Split the last column into 1 and x³; the determinant factors as (Vandermonde)(1+xyz).

   Concept. $\begin{vmatrix}x&x^2&1+x^3\\y&y^2&1+y^3\\z&z^2&1+z^3\end{vmatrix}=\begin{vmatrix}x&x^2&1\\y&y^2&1\\z&z^2&1\end{vmatrix}+\begin{vmatrix}x&x^2&x^3\\y&y^2&y^3\\z&z^2&z^3\end{vmatrix}$.

   Solution. The first determinant, after cycling column 3 to the front (even number of swaps), equals the Vandermonde $V=\begin{vmatrix}1&x&x^2\\1&y&y^2\\1&z&z^2\end{vmatrix}$. The second $=xyz\,V$. So the total is $V(1+xyz)=0$.

   Since $V\neq0$ (given), $1+xyz=0\Rightarrow xyz=-1$.

   ✓Final answer

   (B) $-1$

   ANSWER: (B)

7. GUJCET 2020Set 071 markMCQ
   Q.If $x, y \in R$ and $\begin{vmatrix} (a^x + a^{-x})^2 & (a^x - a^{-x})^2 & 1 \\ (b^x + b^{-x})^2 & (b^x - b^{-x})^2 & 1 \\ (c^x + c^{-x})^2 & (c^x - c^{-x})^2 & 1 \end{vmatrix} = 2y + 6$ then $y =$ ________. (A) $0$ (B) $3$ (C) $-3$ (D) $6$

   ›Reveal solutionSolution

   Column 1 − Column 2 $= 4\times$(Column 3), so the determinant is $0$; $2y+6=0$ gives $y=-3$.

   Concept — spot the linear dependence. For any base $a$, let $p=a^x,\ q=a^{-x}$, so $pq = a^x a^{-x} = 1$. Then

   $(a^x+a^{-x})^2 - (a^x-a^{-x})^2 = 4\,a^x a^{-x} = 4$

   This holds for every row (with bases $a,b,c$). So in the matrix,

   $C_1 - C_2 = 4 = 4\,C_3$

   i.e. $C_1 - C_2 - 4C_3 = \mathbf{0}$ — the columns are linearly dependent, hence

   $\begin{vmatrix}\cdots\end{vmatrix} = 0$

   Given the determinant equals $2y+6$:

   $2y + 6 = 0 \;\Rightarrow\; y = -3$

   ✓Final answer

   Option (C) $-3$

   ANSWER: (C)

8. GUJCET 2021Set 151 markMCQ
   Q.For $\begin{vmatrix} 2 & 3 & 5 \\ 1 & 0 & 7 \\ -1 & -2 & 4 \end{vmatrix}$, the sum of minor and cofactor of $7 = \underline{\hspace{2cm}}$. (A) $0$ (B) $2$ (C) $-2$ (D) $-1$

   ›Reveal solutionSolution

   Element $7$ sits at position $(2,3)$; cofactor $=(-1)^{2+3}\times$ minor.

   Concept: Deleting row 2 and column 3:

   $M_{23}=\begin{vmatrix}2&3\\-1&-2\end{vmatrix}=2(-2)-3(-1)=-1.$

   Cofactor $C_{23}=(-1)^{2+3}M_{23}=-(-1)=1$. Sum $=-1+1=0.$

   ✓Final answer

   (A) $0$

   ANSWER: (A)

9. GUJCET 2024Set 131 markMCQ
   Q.If $\begin{vmatrix} 2017 & 2018 \\ 2019 & 2020 \end{vmatrix} + \begin{vmatrix} 2021 & 2022 \\ 2023 & 2024 \end{vmatrix} = 2k$, then $k^3 =$ __________. (A) $-64$ (B) $-8$ (C) $0$ (D) $8$

   ›Reveal solutionSolution

   Both determinants evaluate to $-2$; their sum $-4=2k$ gives $k=-2$ and $k^3=-8$.

   Concept. Evaluate each $2\times2$ determinant $ad-bc$.

   Steps.

   $\begin{vmatrix}2017&2018\\2019&2020\end{vmatrix}=2017\cdot2020-2018\cdot2019=-2,$

   $\begin{vmatrix}2021&2022\\2023&2024\end{vmatrix}=2021\cdot2024-2022\cdot2023=-2.$

   Sum $=-4=2k\Rightarrow k=-2\Rightarrow k^3=-8.$

   ✓Final answer

   (B) $-8$

   ANSWER: (B)

10. GUJCET 2019Set 171 markMCQ
    Q.Matrix $A_r = \begin{bmatrix} r & r-1 \\ r-1 & r \end{bmatrix}$; $r = 1, 2, 3, \ldots$ If $\sum_{r=1}^{100}|A_r| = (\sqrt{10})^{K}$, then $K = \underline{\hspace{2cm}}$; $(|A_r| = \det(A_r))$. (A) $6$ (B) $4$ (C) $2$ (D) $8$

    ›Reveal solutionSolution

    $|A_r|=r^2-(r-1)^2=2r-1$; the sum of the first 100 odd numbers is $100^2=10000=(\sqrt{10})^8$, so $K=8$.

    Concept: $|A_r|=\begin{vmatrix}r&r-1\r-1&r\end{vmatrix}=r^2-(r-1)^2=2r-1$.

    $\sum_{r=1}^{100}(2r-1)=100^2=10000=10^4=(\sqrt{10})^{8}$

    So $K=8$.

    ✓Final answer

    (D) $8$

    ANSWER: (D)

11. GUJCET 2019Set 171 markMCQ
    Q.If $\begin{vmatrix} 1! & 2! & 3! \\ 2! & 3! & 4! \\ 3! & 4! & 5! \end{vmatrix} = 2016K$, then $K = \underline{\hspace{2cm}}$. (A) $84$ (B) $\dfrac{1}{24}$ (C) $24$ (D) $\dfrac{1}{84}$

    ›Reveal solutionSolution

    Evaluating the determinant of factorials gives $24$; with $24=2016K$, $K=\dfrac{1}{84}$.

    Concept: Write out the values: $1!=1,2!=2,3!=6,4!=24,5!=120$.

    $\begin{vmatrix}1&2&6\\2&6&24\\6&24&120\end{vmatrix}=1(720-576)-2(240-144)+6(48-36)=144-192+72=24$

    Then $24=2016K\Rightarrow K=\dfrac{24}{2016}=\dfrac{1}{84}$.

    ✓Final answer

    (D) $\dfrac{1}{84}$

    ANSWER: (D)

12. GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
    Q.If $f(\theta) = \begin{vmatrix} \cos\theta & -\sin\theta \\ \sin\theta & -\cos\theta \end{vmatrix}$, then $f\left(\frac{\pi}{6}\right) =$ ______.

    (a) $-\frac{1}{2}$

    (b) $\frac{1}{2}$

    (c) $\frac{\sqrt{3}}{2}$

    (d) $-\frac{\sqrt{3}}{2}$

    ›Reveal solutionSolution

    Evaluate the $2\times2$ determinant, simplify with a double-angle identity, then substitute.

    $f(\theta)=\cos\theta(-\cos\theta)-(-\sin\theta)(\sin\theta)=-\cos^2\theta+\sin^2\theta=-\cos2\theta$.

    $f\left(\frac{\pi}{6}\right)=-\cos\frac{\pi}{3}=-\frac{1}{2}$.

    ✓Final answer

    The correct option is (a) $-\frac{1}{2}$.