Q.If $\dfrac{2\sin\alpha}{1 + \cos\alpha + \sin\alpha} = y$, then prove that $\dfrac{1 - \cos\alpha + \sin\alpha}{1 + \sin\alpha}$ is also equal to $y$.

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 $(\sqrt3 - 1)$:

   $\dfrac{(\sqrt3-1)^2}{(\sqrt3+1)(\sqrt3-1)} = \dfrac{3 - 2\sqrt3 + 1}{3-1} = \dfrac{4-2\sqrt3}{2} = 2 - \sqrt3$

   ✓Final answer

   (b) $2 - \sqrt3$.

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 $B = x+y$:

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

   So:

   $\cos(x-y)-\cos(x+y) = -2\sin(x)\sin(-y) = -2\sin x(-\sin y) = 2\sin x\sin y$

   ✓Final answer

   $\cos(x-y)-\cos(x+y) = 2\sin x\sin y$.

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 $\sin^2x+\cos^2x=1$).

   Divide numerator and denominator by $\cos^2 x$: numerator becomes $2\tan x$, denominator becomes $1+\tan^2 x$.

   So $\sin 2x = \dfrac{2\tan x}{1+\tan^2 x}$.

   ✓Final answer

   The correct match is (c) $\dfrac{2\tan x}{1+\tan^2 x}$.

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 $\cos^2x+\sin^2x=1$).

   Divide numerator and denominator by $\cos^2x$: numerator becomes $1-\tan^2x$, denominator becomes $1+\tan^2x$.

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

   ✓Final answer

   The correct match is (a) $\dfrac{1-\tan^2 x}{1+\tan^2 x}$.

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):

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

   ✓Final answer

   The correct match is (b) $\dfrac{2\tan x}{1-\tan^2 x}$.

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

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

   Using $\sin(A+B) = \sin A \cos B + \cos A \sin B$ with $A = B = x$: $\sin(x+x) = \sin x \cos x + \cos x \sin x = 2\sin x \cos x$.

   ✓Final answer

   The correct option is (a) $2\sin x \cos x$.

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 $1 = \cos^2x + \sin^2x = \cos^2x(1+\tan^2x)$):

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

   ✓Final answer

   The correct option is (c) $\dfrac{1-\tan^2 x}{1+\tan^2 x}$.

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

   Rearranging: $4\sin^3 x - 3\sin x = -(3\sin x - 4\sin^3 x) = -\sin 3x$.

   ✓Final answer

   The correct option is (d) $-\sin 3x$.

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

   Taking the reciprocal of both sides: $\cot 2x = \dfrac{1}{\tan 2x} = \dfrac{1-\tan^2 x}{2\tan x}$.

   ✓Final answer

   The correct option is (b) $\cot 2x$.

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).

    It is the standard identity for the cosine of a difference of two angles.

    ✓Final answer

    The correct option is (b) $\cos A \cos B + \sin A \sin B$.

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

    Adding: $\sin(A+B)+\sin(A-B) = (\sin A\cos B + \cos A\sin B) + (\sin A\cos B - \cos A\sin B) = 2\sin A\cos B$ (the $\cos A \sin B$ terms cancel).

    ✓Final answer

    The correct option is (b) $2\sin A \cos B$.

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

    Adding: $\cos(A+B)+\cos(A-B) = (\cos A\cos B - \sin A\sin B) + (\cos A\cos B + \sin A\sin B) = 2\cos A\cos B$ (the $\sin A\sin B$ terms cancel).

    ✓Final answer

    The correct option is (b) $2\cos A \cos B$.