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Exercise 3.3 · Q22

Q.Prove that cot⁡x cot⁡2x−cot⁡2x cot⁡3x−cot⁡3x cot⁡x=1\cot x\, \cot 2x - \cot 2x\, \cot 3x - \cot 3x\, \cot x = 1.

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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⁡θ\sin \theta, cos⁡θ\cos \theta, tan⁡θ\tan \theta) that is true for every angle θ\theta where both sides are defined. It's not a conditional equation (like sin⁡θ=0.5\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⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1

This holds for any angle θ\theta — acute, obtuse, negative, whatever. Why? Because on the unit circle, sin⁡θ\sin \theta is the yy-coordinate and cos⁡θ\cos \theta is the xx-coordinate of a point on a circle of radius 1. The Pythagorean theorem says x2+y2=1x^2 + y^2 = 1, so sin⁡2θ+cos⁡2θ=1\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⁡θ⋅cos⁡θ=sin⁡θ\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⁡θ\tan \theta with sin⁡θcos⁡θ\frac{\sin \theta}{\cos \theta} (a known identity).

tan⁡θ⋅cos⁡θ=sin⁡θcos⁡θ⋅cos⁡θ\tan \theta \cdot \cos \theta = \frac{\sin \theta}{\cos \theta} \cdot \cos \theta

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

=sin⁡θ= \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:

IdentityFormula
Pythagoreansin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1
Quotienttan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta}, cot⁡θ=cos⁡θsin⁡θ\cot \theta = \frac{\cos \theta}{\sin \theta}
Reciprocalcsc⁡θ=1sin⁡θ\csc \theta = \frac{1}{\sin \theta}, sec⁡θ=1cos⁡θ\sec \theta = \frac{1}{\cos \theta}, cot⁡θ=1tan⁡θ\cot \theta = \frac{1}{\tan \theta}
Even-Oddsin⁡(−θ)=−sin⁡θ\sin(-\theta) = -\sin \theta, cos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos \theta
Watch out

A common mistake is to treat sin⁡2θ\sin^2 \theta as (sin⁡θ)2(\sin \theta)^2 — which it is — but then incorrectly think sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 means sin⁡θ+cos⁡θ=1\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: 1−cos⁡2θcos⁡θ=sin⁡θtan⁡θ\frac{1 - \cos^2 \theta}{\cos \theta} = \sin \theta \tan \theta

Start with LHS: 1−cos⁡2θcos⁡θ\frac{1 - \cos^2 \theta}{\cos \theta}

From the Pythagorean identity, 1−cos⁡2θ=sin⁡2θ1 - \cos^2 \theta = \sin^2 \theta. So:

sin⁡2θcos⁡θ=sin⁡θ⋅sin⁡θcos⁡θ=sin⁡θtan⁡θ\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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