Q.Prove that: cos(π/4 - x) cos(π/4 - y) - sin(π/4 - x) sin(π/4 - y) = sin(x + y) OR In a circle of diameter 40 cm, the length of chord is 20 cm. Find the length of minor arc of the chord.

Showing the 12 most recent of 57 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.$\tan 15° = ?$

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

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

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

   (d) $\sqrt{3}$

   ›Reveal solutionSolution

   Write $15° = 45° - 30°$ and apply $\tan(A-B) = \dfrac{\tan A - \tan B}{1+\tan A\tan B}$.

   $\tan 15° = \tan(45° - 30°) = \dfrac{\tan 45° - \tan 30°}{1 + \tan 45°\tan 30°} = \dfrac{1 - \frac{1}{\sqrt3}}{1 + \frac{1}{\sqrt3}} = \dfrac{\sqrt3 - 1}{\sqrt3 + 1}$

   Rationalise by multiplying numerator and denominator by $(\sqrt3 - 1)$: …

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2. CBSE 2026Set ANNUAL1 mark
   Q.cos (x - y) - cos (x + y) = ..............

   ›Reveal solutionSolution

   Use the sum-to-product identity for cos A − cos B to simplify directly.

   Use the identity:

   $\cos A - \cos B = -2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$

   Here $A = x-y$ and $B = x+y$:

   $\frac{A+B}{2} = \frac{(x-y)+(x+y)}{2} = x, \qquad \frac{A-B}{2} = \frac{(x-y)-(x+y)}{2} = -y$

   …

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3. CBSE 2026Set ANNUAL1 markMCQ
   Q.Match the columns — Column A: $\sin 2x$. 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

   $\sin 2x = 2\sin x \cos x$, which in terms of $\tan x$ becomes $\dfrac{2\tan x}{1+\tan^2 x}$.

   Start from $\sin 2x = 2\sin x\cos x$.

   Write $2\sin x\cos x = \dfrac{2\sin x\cos x}{\cos^2 x + \sin^2 x}$ (dividing by 1, since $\sin^2x+\cos^2x=1$).

   …

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4. CBSE 2026Set ANNUAL1 markMCQ
   Q.Match the columns — Column A: $\cos 2x$. 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

   $\cos 2x = \cos^2x - \sin^2x$, which in terms of $\tan x$ becomes $\dfrac{1-\tan^2 x}{1+\tan^2 x}$.

   Start from $\cos 2x = \cos^2x - \sin^2x$.

   Write this as $\dfrac{\cos^2x-\sin^2x}{\cos^2x+\sin^2x}$ (dividing by 1, since $\cos^2x+\sin^2x=1$).

   …

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5. CBSE 2026Set ANNUAL1 markMCQ
   Q.Match the columns — Column A: $\tan 2x$. 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

   $\tan 2x = \dfrac{2\tan x}{1-\tan^2 x}$, obtained by dividing the sine and cosine double-angle identities.

   Using $\sin 2x = \dfrac{2\tan x}{1+\tan^2 x}$ and $\cos 2x = \dfrac{1-\tan^2 x}{1+\tan^2 x}$ (both derived above):

   …

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6. CBSE 2025Set ANNUAL1 markMCQ
   Q.$\sin 2x =$

   (a) $2\sin x \cos x$

   (b) $\sin x \cos x$

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

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

   ›Reveal solutionSolution

   $\sin 2x = 2\sin x \cos x$.

   …

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7. CBSE 2025Set ANNUAL1 markMCQ
   Q.$\cos 2x =$

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

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

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

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

   ›Reveal solutionSolution

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

   Starting from $\cos 2x = \cos^2 x - \sin^2 x$, divide numerator and denominator by $\cos^2 x$ (using $\cos^2x - \sin^2x = \cos^2x(1-\tan^2x)$ and $1 = \cos^2x + \sin^2x = \cos^2x(1+\tan^2x)$):

   …

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8. CBSE 2025Set ANNUAL1 markMCQ
   Q.$4\sin^3 x - 3\sin x =$

   (a) $\sin 3x$

   (b) $\cos 3x$

   (c) $-\cos 3x$

   (d) $-\sin 3x$

   ›Reveal solutionSolution

   $4\sin^3 x - 3\sin x = -\sin 3x$.

   The standard triple-angle identity is $\sin 3x = 3\sin x - 4\sin^3 x$.

   …

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9. CBSE 2025Set ANNUAL1 markMCQ
   Q.$\dfrac{1-\tan^2 x}{2\tan x} =$

   (a) $\tan 2x$

   (b) $\cot 2x$

   (c) $\cos 2x$

   (d) $\sin 2x$

   ›Reveal solutionSolution

   $\dfrac{1-\tan^2 x}{2\tan x} = \cot 2x$.

   The double-angle formula for tangent is $\tan 2x = \dfrac{2\tan x}{1-\tan^2 x}$.

   …

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10. CBSE 2025Set ANNUAL1 markMCQ
    Q.$\cos(A - B) =$

    (a) $\cos A - \cos B$

    (b) $\cos A \cos B + \sin A \sin B$

    (c) $\cos A \cos B - \sin A \sin B$

    (d) $\cos A \sin B - \sin A \cos B$

    ›Reveal solutionSolution

    $\cos(A-B) = \cos A \cos B + \sin A \sin B$.

    This is one of the standard compound angle formulas, derived using the unit circle / geometric construction (or from $\cos(A-B)=\cos(A+(-B))$ and the evenness of cosine, oddness of sine).

    …

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11. CBSE 2025Set ANNUAL1 markMCQ
    Q.$\sin(A + B) + \sin(A - B) =$

    (a) $2\sin A \sin B$

    (b) $2\sin A \cos B$

    (c) $2\cos A \cos B$

    (d) $2\cos A \sin B$

    ›Reveal solutionSolution

    $\sin(A+B) + \sin(A-B) = 2\sin A \cos B$.

    Expand each term: $\sin(A+B) = \sin A\cos B + \cos A\sin B$, and $\sin(A-B) = \sin A\cos B - \cos A\sin B$.

    …

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12. CBSE 2025Set ANNUAL1 markMCQ
    Q.$\cos(A + B) + \cos(A - B) =$

    (a) $2\cos A \sin B$

    (b) $2\cos A \cos B$

    (c) $2\sin A \sin B$

    (d) $2\sin A \cos B$

    ›Reveal solutionSolution

    $\cos(A+B) + \cos(A-B) = 2\cos A \cos B$.

    Expand each term: $\cos(A+B) = \cos A\cos B - \sin A\sin B$, and $\cos(A-B) = \cos A\cos B + \sin A\sin B$.

    …

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