Q.If $\cot\theta = -\dfrac{5}{12}$ and $\theta$ lies in the second quadrant, find the values of $\sin\theta$ and $\sec\theta$.

Showing the 12 most recent of 54 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.The value of $\sin \dfrac{31\pi}{3}$ is

   (a) $\dfrac{\sqrt{3}}{2}$

   (b) $-\dfrac{\sqrt{3}}{2}$

   (c) $\dfrac{1}{2}$

   (d) $\dfrac{1}{\sqrt{2}}$

   ›Reveal solutionSolution

   Subtract multiples of $2\pi$ (period of sine) until the angle lies in $[0, 2\pi)$; $31\pi/3$ reduces to $\pi/3$.

   Since sine has period $2\pi = \dfrac{6\pi}{3}$, subtract multiples of $\dfrac{6\pi}{3}$ from $\dfrac{31\pi}{3}$:

   $\dfrac{31\pi}{3} - 5 \times \dfrac{6\pi}{3} = \dfrac{31\pi}{3} - \dfrac{30\pi}{3} = \dfrac{\pi}{3}$

   So $\sin\dfrac{31\pi}{3} = \sin\dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}$, a standard value.

   ✓Final answer

   (a) $\dfrac{\sqrt{3}}{2}$.

2. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $\tan x = \dfrac{3}{4}$, $\pi < x < \dfrac{3\pi}{2}$ then the value of $\sin \dfrac{x}{2}$ is

   (a) $-\dfrac{1}{\sqrt{10}}$

   (b) $\dfrac{1}{\sqrt{10}}$

   (c) $\dfrac{3}{\sqrt{10}}$

   (d) None of these

   ›Reveal solutionSolution

   Find $\cos x$ from $\tan x$ (both sin, cos negative in Q3), then use the half-angle formula, choosing the sign from the quadrant of $x/2$.

   Given $\tan x = \dfrac{3}{4}$ with $\pi < x < \dfrac{3\pi}{2}$ ($x$ in the third quadrant, where both $\sin x$ and $\cos x$ are negative). Using the 3-4-5 triangle: $\sin x = -\dfrac{3}{5}$, $\cos x = -\dfrac{4}{5}$.

   Since $\pi < x < \dfrac{3\pi}{2}$, dividing by 2: $\dfrac{\pi}{2} < \dfrac{x}{2} < \dfrac{3\pi}{4}$ — so $x/2$ lies in the second quadrant, where sine is positive.

   Half-angle formula: $\sin\dfrac{x}{2} = \pm\sqrt{\dfrac{1-\cos x}{2}} = \sqrt{\dfrac{1-(-4/5)}{2}} = \sqrt{\dfrac{9/5}{2}} = \sqrt{\dfrac{9}{10}} = \dfrac{3}{\sqrt{10}}$

   Taking the positive root (since $x/2$ is in Q2): $\sin\dfrac{x}{2} = \dfrac{3}{\sqrt{10}}$.

   ✓Final answer

   (c) $\dfrac{3}{\sqrt{10}}$.

3. CBSE 2026Set ANNUAL1 markMCQ
   Q.Which one of the following is the value of $\tan \frac{19\pi}{3}$?

   (a) $\sqrt{3}$

   (b) $-\sqrt{3}$

   (c) $\frac{1}{\sqrt{3}}$

   (d) $-\frac{1}{\sqrt{3}}$

   ›Reveal solutionSolution

   $\tan\frac{19\pi}{3} = \sqrt{3}$, option (a).

   Since $\tan\theta$ has period $\pi$, we can subtract integer multiples of $\pi$ from the angle without changing its value.

   $\frac{19\pi}{3} - 6\pi = \frac{19\pi}{3} - \frac{18\pi}{3} = \frac{\pi}{3}$

   So $\tan\frac{19\pi}{3} = \tan\frac{\pi}{3} = \sqrt{3}$.

   ✓Final answer

   The correct option is (a) $\sqrt{3}$.

4. CBSE 2026Set ANNUAL1 markMCQ
   Q.If tan x = -5/12 and x lies in 2nd quadrant, then the value of sin x is:

   (a) 5/13

   (b) 12/13

   (c) -5/13

   (d) -12/13

   ›Reveal solutionSolution

   In the 2nd quadrant sine is positive; using the 5-12-13 right triangle from tan x = -5/12 gives sin x = 5/13.

   Given $\tan x = -\dfrac{5}{12}$ with $x$ in the 2nd quadrant.

   In the 2nd quadrant: $\sin x > 0$, $\cos x < 0$, and $\tan x = \dfrac{\sin x}{\cos x} < 0$ — consistent with the given negative value.

   Treat $5$ and $12$ as the magnitudes of the opposite and adjacent sides; the hypotenuse is $\sqrt{5^2+12^2} = \sqrt{25+144} = \sqrt{169} = 13$.

   Since sine is positive in the 2nd quadrant:

   $\sin x = \frac{5}{13}$

   (Check: $\cos x = -\dfrac{12}{13}$, so $\tan x = \dfrac{5/13}{-12/13} = -\dfrac{5}{12}$ ✓.)

   ✓Final answer

   $\sin x = \dfrac{5}{13}$ — option (a).

5. CBSE 2026Set ANNUAL1 markMCQ
   Q.Assertion (A): The function $\sin x$ is negative in the third and fourth quadrant. Reason (R): The function $\sin x$ is decreasing in the interval $0 \le x \le \dfrac{\pi}{2}$.

   (a) Both Assertion (A) and Reason (R) are correct and Reason (R) is the correct explanation of Assertion (A).

   (b) Both Assertion (A) and Reason (R) are correct, but Reason (R) is not the correct explanation of Assertion (A).

   (c) Assertion (A) is correct, but Reason (R) is incorrect.

   (d) Assertion (A) is incorrect, but Reason (R) is correct.

   ›Reveal solutionSolution

   Assertion (A) is true, but Reason (R) is false — $\sin x$ increases, not decreases, on $[0,\pi/2]$ — so (R) cannot even be a valid explanation of (A).

   Checking Assertion (A): In the third quadrant ($\pi < x < \tfrac{3\pi}{2}$) and fourth quadrant ($\tfrac{3\pi}{2} < x < 2\pi$), the value of $\sin x$ is indeed negative (sine is positive only in the first and second quadrants). So (A) is TRUE.

   Checking Reason (R): On the interval $0\le x\le \tfrac{\pi}{2}$, $\sin x$ rises from $\sin 0 = 0$ to $\sin(\pi/2)=1$ — it is strictly INCREASING on this interval, not decreasing. So (R) is FALSE.

   Since (A) is true and (R) is false, (R) also cannot be a correct explanation of (A).

   ✓Final answer

   Option (c): Assertion (A) is correct, but Reason (R) is incorrect.

6. CBSE 2026Set ANNUAL1 markMCQ
   Q.Match the columns — Column A: $\sin(-x)$. Choose its correct equivalent from Column B.

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

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

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

   (d) $-\sin x$

   (e) $\sin x$

   (f) $-\cos x$

   (g) $\cos x$

   ›Reveal solutionSolution

   Sine is an odd function: $\sin(-x)=-\sin x$ for all $x$.

   This follows directly from the standard trigonometric identity for negative angles, provable from the unit-circle definition: reflecting the angle across the x-axis flips the sign of the y-coordinate (sine) while leaving the x-coordinate (cosine) unchanged.

   ✓Final answer

   The correct match is (d) $-\sin x$.

7. CBSE 2026Set ANNUAL1 markMCQ
   Q.Match the columns — Column A: $\cos\left(\dfrac{\pi}{2} - x\right)$. Choose its correct equivalent from Column B.

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

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

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

   (d) $-\sin x$

   (e) $\sin x$

   (f) $-\cos x$

   (g) $\cos x$

   ›Reveal solutionSolution

   By the co-function identity, $\cos\left(\dfrac{\pi}{2}-x\right)=\sin x$.

   This is a standard complementary-angle identity: the cosine of $90^\circ$ (i.e. $\tfrac{\pi}{2}$) minus an angle equals the sine of that angle, following from the right-triangle definitions where the two acute angles are complementary.

   ✓Final answer

   The correct match is (e) $\sin x$.

8. CBSE 2026Set ANNUAL1 markMCQ
   Q.Match the columns — Column A: $\cos(\pi - x)$. Choose its correct equivalent from Column B.

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

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

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

   (d) $-\sin x$

   (e) $\sin x$

   (f) $-\cos x$

   (g) $\cos x$

   ›Reveal solutionSolution

   By the supplementary-angle identity, $\cos(\pi-x)=-\cos x$.

   This follows since $\pi-x$ lies in the second quadrant when $x$ is a first-quadrant angle, where cosine is negative, and the reference angle is $x$ itself, giving magnitude $\cos x$ with a negative sign.

   ✓Final answer

   The correct match is (f) $-\cos x$.

9. CBSE 2026Set 1A1 mark
   Q.Find the value of the $\sin\left(-\frac{11\pi}{3}\right)$.

   ›Reveal solutionSolution

   $-\tfrac{11\pi}{3}+4\pi=\tfrac{\pi}{3}$, so the value is $\sin\tfrac{\pi}{3}=\tfrac{\sqrt3}{2}$.

   Using periodicity $\sin\theta=\sin(\theta+2\pi k)$:

   $-\frac{11\pi}{3}+4\pi=-\frac{11\pi}{3}+\frac{12\pi}{3}=\frac{\pi}{3}.$

   Hence $\sin\left(-\dfrac{11\pi}{3}\right)=\sin\dfrac{\pi}{3}=\dfrac{\sqrt3}{2}.$

   ✓Final answer

   $\sin\left(-\dfrac{11\pi}{3}\right)=\dfrac{\sqrt3}{2}.$

10. CBSE 2025Set ANNUAL1 markMCQ
    Q.$\sin \dfrac{3\pi}{2} =$

    (a) 0

    (b) -1

    (c) 1

    (d) 1/2

    ›Reveal solutionSolution

    $\sin\dfrac{3\pi}{2} = -1$.

    $\dfrac{3\pi}{2}$ radians $= 270°$. On the unit circle, $270°$ corresponds to the point $(0, -1)$ on the negative $y$-axis, where $\sin\theta$ equals the $y$-coordinate, i.e., $-1$.

    ✓Final answer

    The correct option is (b) $-1$.

11. CBSE 2025Set ANNUAL1 markMCQ
    Q.$\sin(\pi + x) =$

    (a) $\cos x$

    (b) $-\cos x$

    (c) $\sin x$

    (d) $-\sin x$

    ›Reveal solutionSolution

    $\sin(\pi + x) = -\sin x$.

    Using the compound angle formula $\sin(\pi + x) = \sin\pi\cos x + \cos\pi\sin x$. Since $\sin\pi = 0$ and $\cos\pi = -1$, this gives $\sin(\pi+x) = 0 \cdot \cos x + (-1)\sin x = -\sin x$.

    ✓Final answer

    The correct option is (d) $-\sin x$.

12. CBSE 2025Set ANNUAL1 markMCQ
    Q.$\cos \dfrac{5\pi}{2} =$

    (a) 0

    (b) 1

    (c) -1

    (d) 1/2

    ›Reveal solutionSolution

    $\cos\dfrac{5\pi}{2} = 0$.

    Since cosine has period $2\pi$: $\dfrac{5\pi}{2} - 2\pi = \dfrac{5\pi}{2} - \dfrac{4\pi}{2} = \dfrac{\pi}{2}$.

    So $\cos\dfrac{5\pi}{2} = \cos\dfrac{\pi}{2} = 0$.

    ✓Final answer

    The correct option is (a) 0.