$\sin 10^\circ$ is greater than $\cos 10^\circ$.

sin 10^° is greater than cos 10^°.

NCERT Exemplar · Q70

\sin 10^\circ

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Concept understanding — Trigonometric Identity Proof

Trigonometric Identity Proof: From Intuition to Precision

Imagine you're standing at the corner of a right triangle. The two shorter sides — one horizontal, one vertical — and the sloping hypotenuse are all connected. If you change the angle at your corner, the lengths of the sides change, but the relationship between them stays fixed. That fixed relationship is what a trigonometric identity captures.

The Core Idea

A trigonometric identity is an equation involving trigonometric functions (like $\sin \theta$ , $\cos \theta$ , $\tan \theta$ ) that is true for every angle $\theta$ where both sides are defined. It's not a conditional equation (like $\sin \theta = 0.5$ , which is true only for specific angles). It's an eternal truth about how these functions relate.

The most famous one is:

\sin^2 \theta + \cos^2 \theta = 1

This holds for any angle $\theta$ — acute, obtuse, negative, whatever. Why? Because on the unit circle, $\sin \theta$ is the $y$ -coordinate and $\cos \theta$ is the $x$ -coordinate of a point on a circle of radius 1. The Pythagorean theorem says $x^2 + y^2 = 1$ , so $\sin^2 \theta + \cos^2 \theta = 1$ is just the Pythagorean theorem in disguise.

Proving an Identity: The Method

When you're asked to prove a trigonometric identity, you're not solving for an angle. You're showing that the left-hand side (LHS) and right-hand side (RHS) are the same expression, just written differently.

The golden rule: Start with one side and transform it into the other, using known identities and algebraic manipulation. Never move terms across the equals sign as if solving an equation — that assumes the identity is already true, which is what you're trying to prove.

A Simple Example

Prove: $\tan \theta \cdot \cos \theta = \sin \theta$

Step 1: Pick a side to start with. Usually, the more complicated side is easier to simplify. Here, the LHS looks more complex.

Step 2: Replace $\tan \theta$ with $\frac{\sin \theta}{\cos \theta}$ (a known identity).

\tan \theta \cdot \cos \theta = \frac{\sin \theta}{\cos \theta} \cdot \cos \theta

Step 3: Cancel $\cos \theta$ (provided $\cos \theta \neq 0$ — but the identity holds for all angles where both sides are defined, and at $\cos \theta = 0$ , $\tan \theta$ is undefined anyway).

= \sin \theta

That's it. The LHS simplifies exactly to the RHS.

The Toolbox of Known Identities

To prove any identity, you need to know the basic building blocks:

Identity	Formula
Pythagorean	$\sin^2 \theta + \cos^2 \theta = 1$
Quotient	$\tan \theta = \frac{\sin \theta}{\cos \theta}$ , $\cot \theta = \frac{\cos \theta}{\sin \theta}$
Reciprocal	$\csc \theta = \frac{1}{\sin \theta}$ , $\sec \theta = \frac{1}{\cos \theta}$ , $\cot \theta = \frac{1}{\tan \theta}$
Even-Odd	$\sin(-\theta) = -\sin \theta$ , $\cos(-\theta) = \cos \theta$

Watch out

A common mistake is to treat $\sin^2 \theta$ as $(\sin \theta)^2$ — which it is — but then incorrectly think $\sin^2 \theta + \cos^2 \theta = 1$ means $\sin \theta + \cos \theta = 1$ . It does not. The square applies to the whole sine value, not to the angle.

A Slightly Harder Proof

Prove: $\frac{1 - \cos^2 \theta}{\cos \theta} = \sin \theta \tan \theta$

Start with LHS: $\frac{1 - \cos^2 \theta}{\cos \theta}$

From the Pythagorean identity, $1 - \cos^2 \theta = \sin^2 \theta$ . So:

\frac{\sin^2 \theta}{\cos \theta} = \sin \theta \cdot \frac{\sin \theta}{\cos \theta} = \sin \theta \tan \theta

That's the RHS. Done. …

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Compare $\sin 10°$ and $\cos 10°$ using the complementary angle relationship: since $10° < 45°$ , we have $\sin 10° < \cos 10°$ . The statement is false.

The key insight here rests on understanding how sine and cosine behave in the first quadrant and their complementary relationship.

For any angle $\theta$ in the first quadrant ( $0° < \theta < 90°$ ), we know that $\sin \theta = \cos(90° - \theta)$ . This means:

\sin 10° = \cos 80°

Now we need to compare $\cos 80°$ with $\cos 10°$ . The cosine function is decreasing on the interval $[0°, 90°]$ —as the angle increases from $0°$ to $90°$ , cosine decreases from $1$ to $0$ .

Let me walk through the comparison:

Apply the complementary angle identity

We have $\sin 10° = \cos(90° - 10°) = \cos 80°$ .
Use the monotonicity of cosine

Since cosine decreases as the angle increases in the first quadrant, and $80° > 10°$ , we get:

\cos 80° < \cos 10°

Substitute back From step 1, this means: $\sin 10° < \cos 10°$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 57 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$\tan 15° = ?$
(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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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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CBSE 2026Set ANNUAL1 markMCQ
$\sin 2x$
(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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CBSE 2026Set ANNUAL1 markMCQ
$\cos 2x$
(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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CBSE 2026Set ANNUAL1 markMCQ
$\tan 2x$
(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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CBSE 2025Set ANNUAL1 markMCQ
$\sin 2x =$
(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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CBSE 2025Set ANNUAL1 markMCQ
$\cos 2x =$
(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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CBSE 2025Set ANNUAL1 markMCQ
$4\sin^3 x - 3\sin x =$
(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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CBSE 2025Set ANNUAL1 markMCQ
$\dfrac{1-\tan^2 x}{2\tan x} =$
(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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CBSE 2025Set ANNUAL1 markMCQ
$\cos(A - B) =$
(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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CBSE 2025Set ANNUAL1 markMCQ
$\sin(A + B) + \sin(A - B) =$
(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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CBSE 2025Set ANNUAL1 markMCQ
$\cos(A + B) + \cos(A - B) =$
(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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Q.sin⁡10∘\sin 10^\circsin10∘ is greater than cos⁡10∘\cos 10^\circcos10∘.

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Q. $\sin 10^\circ$ is greater than $\cos 10^\circ$ .

Unlock everything free for 14 days