Q.One of the values of $x$ for which $\begin{vmatrix} \cos x & \sin x \\ -\cos x & \sin x \end{vmatrix} = 1$ is
(A) $0$
(B) $\dfrac{\pi}{4}$
(C) $\dfrac{\pi}{3}$
(D) $\dfrac{\pi}{2}$

Showing the 12 most recent of 63 on this concept.

1. CBSE 2026Set 65/1/11 markMCQ
   Q.One of the values of $x$ for which $\begin{vmatrix} \cos x & \sin x \\ -\cos x & \sin x \end{vmatrix} = 1$ is (A) $0$ (B) $\dfrac{\pi}{4}$ (C) $\dfrac{\pi}{3}$ (D) $\dfrac{\pi}{2}$

   ›Reveal solutionSolution

   The determinant simplifies to $\sin x \cos x + \sin x \cos x = \sin 2x$. Setting $\sin 2x = 1$ gives $2x = \frac{\pi}{2} + 2n\pi$, so $x = \frac{\pi}{4}$ is one solution. The correct option is (B).

   The problem gives a $2 \times 2$ determinant equal to $1$ and asks for a value of $x$ from the options. The fastest route is to compute the determinant directly — it’s a simple expression in $\sin x$ and $\cos x$ — and then solve the resulting trigonometric equation.

   The determinant of $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$ is $ad - bc$. Here:

   $\begin{vmatrix} \cos x & \sin x \\ -\cos x & \sin x \end{vmatrix} = (\cos x)(\sin x) - (\sin x)(-\cos x)$

   1. Simplify the expression. The first term is $\cos x \sin x$. The second term: $(\sin x)(-\cos x) = -\sin x \cos x$, but there’s a minus sign in front, so it becomes $-(-\sin x \cos x) = +\sin x \cos x$. So the determinant equals:

   $\cos x \sin x + \sin x \cos x = 2 \sin x \cos x$

   2. Use the double-angle identity. Recall that $2 \sin x \cos x = \sin 2x$. Therefore the equation becomes:

   $\sin 2x = 1$

   3. Solve $\sin 2x = 1$. The sine function equals $1$ at $\frac{\pi}{2}$ plus any integer multiple of $2\pi$:

   $2x = \frac{\pi}{2} + 2n\pi \quad \Rightarrow \quad x = \frac{\pi}{4} + n\pi$

   where $n$ is any integer.

   4. Check the given options.
      • (A) $0$: $\sin 0 = 0$, not $1$.
      • (B) $\frac{\pi}{4}$: $\sin \frac{\pi}{2} = 1$ — works.
      • (C) $\frac{\pi}{3}$: $\sin \frac{2\pi}{3} = \frac{\sqrt{3}}{2}$, not $1$.
      • (D) $\frac{\pi}{2}$: $\sin \pi = 0$, not $1$.

   Watch out

   A common mistake is to forget the minus sign in the determinant expansion. Here, $ad - bc$ with $b = \sin x$ and $c = -\cos x$ gives $(\cos x)(\sin x) - (\sin x)(-\cos x) = \cos x \sin x + \sin x \cos x$. If you miss the double negative, you’d get $0$ and no solution.

   Tip

   Recognizing $2 \sin x \cos x = \sin 2x$ immediately turns a determinant problem into a basic trigonometric equation — always look for double-angle forms when you see products of sine and cosine.

   ✓Final answer

   The correct option is (B) $\dfrac{\pi}{4}$.

2. CBSE 2026Set CX1 markMCQ
   Q.$\sin(\tan^{-1}x)$, $|x|<1$ is equal to:

   (a) $\dfrac{x}{\sqrt{1+x^{2}}}$

   (b) $\dfrac{x}{\sqrt{1-x^{2}}}$

   (c) $\dfrac{1}{\sqrt{1+x^{2}}}$

   (d) $\dfrac{1}{\sqrt{1-x^{2}}}$

   ›Reveal solutionSolution

   With $\theta=\tan^{-1}x$ a right triangle gives $\sin\theta=\dfrac{x}{\sqrt{1+x^{2}}}$ — option (a).

   Concept: Convert the inverse function to an angle and read the ratio off a right triangle.

   Let $\theta=\tan^{-1}x$, so $\tan\theta=x$. Take the opposite side $=x$ and adjacent $=1$; then the hypotenuse is $\sqrt{1+x^{2}}$.

   $\sin\theta=\dfrac{\text{opposite}}{\text{hypotenuse}}=\dfrac{x}{\sqrt{1+x^{2}}}.$

   ✓Final answer

   Option (a) $\;\sin(\tan^{-1}x)=\dfrac{x}{\sqrt{1+x^{2}}}$.

3. CBSE 2026Set A1 markMCQ
   Q.$\sin(\cos^{-1}3/5) =$

   (a) $\frac{3}{4}$

   (b) $\frac{4}{5}$

   (c) $\frac{3}{5}$

   (d) $\frac{5}{4}$

   ›Reveal solutionSolution

   $\sin\left(\cos^{-1}\frac{3}{5}\right) = \frac{4}{5}$.

   Let $\theta = \cos^{-1}\frac{3}{5}$, so $\cos\theta = \frac{3}{5}$ with $\theta\in[0,\pi]$, where $\sin\theta \ge 0$.

   Then

   $\sin\theta = \sqrt{1-\cos^2\theta} = \sqrt{1-\frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}.$

   ✓Final answer

   (b) $\frac{4}{5}$.

4. CBSE 2026Set A1 markMCQ
   Q.If $|x| \le 1$, then $\tan(\cos^{-1}x) =$

   (a) $\frac{\sqrt{1-x^2}}{x}$

   (b) $\frac{x}{1+x^2}$

   (c) $\frac{\sqrt{1+x^2}}{x}$

   (d) $\sqrt{1-x^2}$

   ›Reveal solutionSolution

   $\tan(\cos^{-1}x) = \frac{\sqrt{1-x^2}}{x}$.

   Let $\theta = \cos^{-1}x$, so $\cos\theta = x$ with $\theta\in[0,\pi]$ (where $\sin\theta\ge 0$).

   Then $\sin\theta = \sqrt{1-x^2}$, and

   $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\sqrt{1-x^2}}{x}.$

   ✓Final answer

   (a) $\frac{\sqrt{1-x^2}}{x}$.

5. CBSE 2026Set A1 markMCQ
   Q.$\int\frac{\cos 2x}{(\sin x + \cos x)^2}\,dx =$

   (a) $2\log(\sin x + \cos x) + k$

   (b) $\log(\sin x + \cos x) + k$

   (c) $\log(\sin x - \cos x) + k$

   (d) $-\frac{1}{\sin x + \cos x} + k$

   ›Reveal solutionSolution

   Factor $\cos 2x$ and substitute $u=\sin x+\cos x$ to get $\log(\sin x+\cos x)+k$.

   Write $\cos 2x=\cos^2 x-\sin^2 x=(\cos x-\sin x)(\cos x+\sin x)$. Then

   $\dfrac{\cos 2x}{(\sin x+\cos x)^2}=\dfrac{(\cos x-\sin x)(\cos x+\sin x)}{(\sin x+\cos x)^2}=\dfrac{\cos x-\sin x}{\sin x+\cos x}$.

   Let $u=\sin x+\cos x$, so $du=(\cos x-\sin x)\,dx$. The integral becomes

   $\int\dfrac{du}{u}=\log|u|+k=\log(\sin x+\cos x)+k$.

   ✓Final answer

   (B) $\log(\sin x+\cos x)+k$.

6. CBSE 2026Set A1 markMCQ
   Q.$\int\frac{1 - \cos 2x}{1 + \cos 2x}\,dx =$

   (a) $\tan x + x + k$

   (b) $\tan x - x + k$

   (c) $x - \tan^2 x + k$

   (d) $\tan\frac{x}{2} + k$

   ›Reveal solutionSolution

   Simplify to $\tan^2 x$, then integrate: $\tan x - x + k$.

   Use $1-\cos 2x=2\sin^2 x$ and $1+\cos 2x=2\cos^2 x$:

   $\dfrac{1-\cos 2x}{1+\cos 2x}=\dfrac{2\sin^2 x}{2\cos^2 x}=\tan^2 x=\sec^2 x-1$.

   Therefore $\int\tan^2 x\,dx=\int(\sec^2 x-1)\,dx=\tan x-x+k$.

   ✓Final answer

   (B) $\tan x - x + k$.

7. CBSE 2026Set ANNUAL1 markMCQ
   Q.$\int \frac{dx}{\sin^2 x \cos^2 x}$ equals

   (a) $\tan x + \sin x + c$

   (b) $\tan x - \cot x + c$

   (c) $\tan x \cot x + c$

   (d) $2\tan x - \cot 2x + c$

   ›Reveal solutionSolution

   Split the integrand using $\sin^2x+\cos^2x=1$ in the numerator, then integrate each standard term.

   $\dfrac{1}{\sin^2x\cos^2x} = \dfrac{\sin^2x+\cos^2x}{\sin^2x\cos^2x} = \dfrac{1}{\cos^2x}+\dfrac{1}{\sin^2x} = \sec^2x+\csc^2x$.

   $\displaystyle\int(\sec^2x+\csc^2x)\,dx = \tan x - \cot x + c$.

   ✓Final answer

   The correct option is (b) $\tan x-\cot x+c$.

8. CBSE 2026Set ANNUAL1 markMCQ
   Q.$\int \tan^2 x \, dx =$

   (a) $\cot x - x + C$

   (b) $\tan x + x + C$

   (c) $\tan x - x + C$

   (d) None of these

   ›Reveal solutionSolution

   Rewrite $\tan^2 x$ using the identity $\tan^2 x = \sec^2 x - 1$, then integrate term by term.

   $\int \tan^2 x\,dx = \int(\sec^2 x - 1)\,dx = \tan x - x + C$.

   ✓Final answer

   (c) $\tan x - x + C$.

9. CBSE 2026Set ANNUAL1 mark
   Q.If tan⁻¹(1/3) = x, then find sin x.

   ›Reveal solutionSolution

   Build a right triangle using $\tan x = 1/3$ and read off $\sin x$.

   Given $\tan^{-1}(1/3)=x \Rightarrow \tan x = 1/3$.

   In a right triangle take opposite side $=1$, adjacent side $=3$, so hypotenuse $=\sqrt{1^2+3^2}=\sqrt{10}$.

   $\sin x = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{1}{\sqrt{10}}$

   ✓Final answer

   $\sin x = \dfrac{1}{\sqrt{10}}$.

10. CBSE 2026Set ANNUAL1 markMCQ
    Q.The value of \int \frac{sec^2 x}{cosec^2 x} dx is:

    (a)

    (i) tan x - x + c

    (b)

    (ii) tan x + x + c

    (c)

    (iii) cot x - x + c

    (d)

    (iv) log cosec x + c

    ›Reveal solutionSolution

    $\displaystyle\int \dfrac{\sec^2 x}{\csc^2 x}\,dx = \tan x - x + c$ — option (i).

    Concept. Convert to a single trigonometric ratio, then use the identity $\tan^2 x=\sec^2 x-1$ and the standard integral $\int\sec^2 x\,dx=\tan x$.

    Steps.

    • $\dfrac{\sec^2 x}{\csc^2 x}=\dfrac{1/\cos^2 x}{1/\sin^2 x}=\dfrac{\sin^2 x}{\cos^2 x}=\tan^2 x$.

    • Rewrite: $\tan^2 x=\sec^2 x-1$.

    • Integrate: $\displaystyle\int(\sec^2 x-1)\,dx=\tan x - x + c$.

    ✓Final answer

    $\tan x - x + c$ — option (i).

11. CBSE 2026Set ANNUAL1 mark
    Q.Evaluate: $\sin\left\{\cos^{-1}\left(-\dfrac{4}{5}\right)\right\}$

    ›Reveal solutionSolution

    With $\cos\theta=-\tfrac45$ and $\theta\in[0,\pi]$, $\sin\theta=+\tfrac35$.

    Let $\theta=\cos^{-1}\left(-\tfrac45\right)$, so $\cos\theta=-\tfrac45$ and $\theta\in[0,\pi]$ (range of $\cos^{-1}$). On this range $\sin\theta\ge 0$.

    $\sin\theta=\sqrt{1-\cos^{2}\theta}=\sqrt{1-\tfrac{16}{25}}=\sqrt{\tfrac{9}{25}}=\tfrac{3}{5}.$

    ✓Final answer

    $\dfrac{3}{5}$.

12. CBSE 2025Set 65/4/11 markMCQ
    Q.$\int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha}\, dx$ is equal to : (A) $2(\sin x + x \cos \alpha) + C$ (B) $2(\sin x - x \cos \alpha) + C$ (C) $2(\sin x + 2x \cos \alpha) + C$ (D) $2(\sin x + \sin \alpha) + C$

    ›Reveal solutionSolution

    Use the cosine difference identity to simplify the numerator, then factor and cancel the denominator. The integral reduces to $2(\sin x + x \cos \alpha) + C$, matching option (A).

    The key here is to recognise that the integrand looks messy, but the numerator and denominator are both differences of cosines. That structure is a direct invitation to use the identity:

    $\cos A - \cos B = -2 \sin\frac{A+B}{2} \sin\frac{A-B}{2}$

    Applying this to both the numerator and denominator will let us cancel common factors and turn the integral into something elementary.

    ---

    1. Rewrite the numerator using the identity above, with $A = 2x$ and $B = 2\alpha$:

    $\cos 2x - \cos 2\alpha = -2 \sin\frac{2x+2\alpha}{2} \sin\frac{2x-2\alpha}{2} = -2 \sin(x+\alpha) \sin(x-\alpha)$

    2. Rewrite the denominator similarly, with $A = x$ and $B = \alpha$:

    $\cos x - \cos \alpha = -2 \sin\frac{x+\alpha}{2} \sin\frac{x-\alpha}{2}$

    3. Form the integrand by dividing the two expressions. The minus signs cancel:

    $\frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} = \frac{-2 \sin(x+\alpha) \sin(x-\alpha)}{-2 \sin\frac{x+\alpha}{2} \sin\frac{x-\alpha}{2}} = \frac{\sin(x+\alpha) \sin(x-\alpha)}{\sin\frac{x+\alpha}{2} \sin\frac{x-\alpha}{2}}$

    4. Use the double-angle identity for sine: $\sin \theta = 2 \sin\frac{\theta}{2} \cos\frac{\theta}{2}$. Apply it to both factors in the numerator:

    $\sin(x+\alpha) = 2 \sin\frac{x+\alpha}{2} \cos\frac{x+\alpha}{2}$

    $\sin(x-\alpha) = 2 \sin\frac{x-\alpha}{2} \cos\frac{x-\alpha}{2}$

    Substitute these into the fraction:

    $\frac{ \left(2 \sin\frac{x+\alpha}{2} \cos\frac{x+\alpha}{2}\right) \left(2 \sin\frac{x-\alpha}{2} \cos\frac{x-\alpha}{2}\right) } { \sin\frac{x+\alpha}{2} \sin\frac{x-\alpha}{2} }$

    The $\sin$ terms cancel completely, leaving:

    $4 \cos\frac{x+\alpha}{2} \cos\frac{x-\alpha}{2}$

    5. Simplify the product of cosines using the identity:

    $\cos P \cos Q = \frac{1}{2} \left[ \cos(P+Q) + \cos(P-Q) \right]$

    Here $P = \frac{x+\alpha}{2}$ and $Q = \frac{x-\alpha}{2}$. Then:

    $P+Q = \frac{x+\alpha + x - \alpha}{2} = x$

    $P-Q = \frac{x+\alpha - (x-\alpha)}{2} = \alpha$

    So:

    $4 \cdot \frac{1}{2} \left[ \cos x + \cos \alpha \right] = 2(\cos x + \cos \alpha)$

    The integrand is now simply $2(\cos x + \cos \alpha)$.

    Tip

    The whole messy expression collapses to a sum of two cosines — one depending on $x$, one constant. That’s the cleanest possible form for integration.

    6. Integrate term by term:

    $\int 2(\cos x + \cos \alpha) \, dx = 2 \int \cos x \, dx + 2 \cos \alpha \int 1 \, dx$

    $= 2 \sin x + 2x \cos \alpha + C$

    Watch out

    A common mistake is to forget that $\cos \alpha$ is a constant with respect to $x$, so its integral is $x \cos \alpha$, not $\sin \alpha$ or something else.

    ---

    ✓Final answer

    The integral equals $2(\sin x + x \cos \alpha) + C$, which corresponds to option (A).