Q.Prove that: $\tan x \tan 2x \tan 3x = \tan 3x - \tan 2x - \tan x$ OR Prove that: $\cos\left(\dfrac{\pi}{4} + x\right) + \cos\left(\dfrac{\pi}{4} - x\right) = \sqrt{2} \cos x$

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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