If $m\sin\theta = n\sin(\theta + 2\alpha)$, then prove that $\tan(\theta + \alpha)\cot\alpha = \dfrac{m + n}{m - n}$.

If msinθ = nsin(θ + 2α), then prove that tan(θ + α – Class 11 Mathematics

NCERT Exemplar · Q3

m\sin\theta = n\sin(\theta + 2\alpha)

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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.

What Makes a Proof Valid?

Every step must be reversible or an equivalence. You're not solving; you're rewriting.
State any restrictions. If you divide by $\cos \theta$ , note that $\cos \theta \neq 0$ for that step — but the identity may still hold in the limit.
Work on one side only. The cleanest proofs transform LHS into RHS (or vice versa) without touching both sides simultaneously.

Tip

If you get stuck, try rewriting everything in terms of $\sin \theta$ and $\cos \theta$ . Most identities become simple algebra after that.

The Big Picture

Trigonometric identities are the grammar of trigonometry. They let you simplify complex expressions, solve equations, and later integrate trigonometric functions in calculus. Every proof is just a puzzle: "Can I connect these two expressions using the relationships I already know?"

Start with the simplest identity — $\sin^2 \theta + \cos^2 \theta = 1$ — and build from there. With practice, you'll see the patterns: factor, substitute, cancel, rewrite. That's all there is to it.

Proving trigonometric identities using the Pythagorean, quotient, and reciprocal relations is a staple exercise in the NCERT Class 11 Mathematics chapter on Trigonometric Functions, and "how to prove trigonometric identities step by step" is a commonly searched topic for CBSE board and JEE Main preparation. Because these identities are reused throughout calculus and coordinate geometry, they are consistently featured in "trigonometric identities important questions" for competitive-exam practice.

The key idea is to use the Componendo and Dividendo rule along with sum-to-product trigonometric identities.

Rearrange the given equation $m\sin\theta = n\sin(\theta + 2\alpha)$ to form a ratio:

\frac{\sin(\theta + 2\alpha)}{\sin\theta} = \frac{m}{n}

Apply the Componendo and Dividendo rule to both sides:

\frac{\sin(\theta + 2\alpha) + \sin\theta}{\sin(\theta + 2\alpha) - \sin\theta} = \frac{m + n}{m - n}

Use the sum-to-product formulas $\sin A + \sin B = 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$ and $\sin A - \sin B = 2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$ on the left-hand side. Here, $A = \theta + 2\alpha$ and $B = \theta$ :

\frac{2\sin\left(\frac{\theta + 2\alpha + \theta}{2}\right)\cos\left(\frac{\theta + 2\alpha - \theta}{2}\right)}{2\cos\left(\frac{\theta + 2\alpha + \theta}{2}\right)\sin\left(\frac{\theta + 2\alpha - \theta}{2}\right)} = \frac{m + n}{m - n}

Simplify the expression:

\frac{2\sin(\theta + \alpha)\cos\alpha}{2\cos(\theta + \alpha)\sin\alpha} = \frac{m + n}{m - n}

\frac{\sin(\theta + \alpha)}{\cos(\theta + \alpha)} \cdot \frac{\cos\alpha}{\sin\alpha} = \frac{m + n}{m - n}

\tan(\theta + \alpha)\cot\alpha = \frac{m + n}{m - n}

✓Final answer

\tan(\theta + \alpha)\cot\alpha = \boxed{\dfrac{m + n}{m - n}}

The problem is solved by applying the Componendo and Dividendo rule to the given ratio, followed by using sum-to-product trigonometric identities, to prove that $\tan(\theta + \alpha)\cot\alpha = \dfrac{m + n}{m - n}$ .

This problem asks us to prove a trigonometric identity. The key to approaching such proofs is often to look at the structure of the given equation and the expression we need to prove.

Notice the term $\dfrac{m+n}{m-n}$ on the right-hand side of the expression we need to prove. This specific form, a sum divided by a difference, is a strong indicator that the Componendo and Dividendo rule will be useful. This rule allows us to transform a ratio $\frac{a}{b} = \frac{c}{d}$ into $\frac{a+b}{a-b} = \frac{c+d}{c-d}$ , which perfectly matches the structure of the $m$ and $n$ terms.

On the left-hand side of the target expression, we have $\tan(\theta + \alpha)\cot\alpha$ . This involves angles $\theta + \alpha$ and $\alpha$ . The given equation has $\sin\theta$ and $\sin(\theta + 2\alpha)$ . If we apply Componendo and Dividendo to the sines, we will get sums and differences of sines, which can then be converted into products using the sum-to-product formulas. These formulas are designed to simplify expressions like $\sin A \pm \sin B$ into products involving half-angles, which often align with the angles we need.

Let's proceed with the proof step-by-step.

Rearrange the given equation into a ratio. The given equation is $m\sin\theta = n\sin(\theta + 2\alpha)$ . To apply Componendo and Dividendo, we need to express this as a ratio. We can write:

\frac{\sin(\theta + 2\alpha)}{\sin\theta} = \frac{m}{n}

Apply the Componendo and Dividendo rule. This rule states that if $\frac{a}{b} = \frac{c}{d}$ , then $\frac{a+b}{a-b} = \frac{c+d}{c-d}$ . Applying this to our ratio:

\frac{\sin(\theta + 2\alpha) + \sin\theta}{\sin(\theta + 2\alpha) - \sin\theta} = \frac{m + n}{m - n}

This step directly gives us the right-hand side of the expression we need to prove. Now we need to simplify the left-hand side.

> [!FORMULA] Componendo and Dividendo Rule
> If $\frac{a}{b} = \frac{c}{d}$, then $\frac{a+b}{a-b} = \frac{c+d}{c-d}$.

3. Use sum-to-product formulas on the left-hand side.

The numerator and denominator on the LHS are in the form $\sin A + \sin B$ and $\sin A - \sin B$ .

Let $A = \theta + 2\alpha$ and $B = \theta$ .

We need to calculate $\frac{A+B}{2}$ and $\frac{A-B}{2}$ :

\frac{A+B}{2} = \frac{(\theta + 2\alpha) + \theta}{2} = \frac{2\theta + 2\alpha}{2} = \theta + \alpha

\frac{A-B}{2} = \frac{(\theta + 2\alpha) - \theta}{2} = \frac{2\alpha}{2} = \alpha

Now, apply the sum-to-product identities:
> [!FORMULA] Sum-to-Product Identities
> $\sin A + \sin B = 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$
> $\sin A - \sin B = 2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$

Substitute $A = \theta + 2\alpha$ and $B = \theta$ into these formulas:
*   Numerator: $\sin(\theta + 2\alpha) + \sin\theta = 2\sin(\theta + \alpha)\cos\alpha$
*   Denominator: $\sin(\theta + 2\alpha) - \sin\theta = 2\cos(\theta + \alpha)\sin\alpha$

4. Substitute these back into the LHS and simplify.

The left-hand side becomes:

\frac{2\sin(\theta + \alpha)\cos\alpha}{2\cos(\theta + \alpha)\sin\alpha}

Cancel out the $2$s:

\frac{\sin(\theta + \alpha)\cos\alpha}{\cos(\theta + \alpha)\sin\alpha}

We can rearrange this as a product of two ratios:

\left(\frac{\sin(\theta + \alpha)}{\cos(\theta + \alpha)}\right) \cdot \left(\frac{\cos\alpha}{\sin\alpha}\right)

Recall that $\frac{\sin x}{\cos x} = \tan x$ and $\frac{\cos x}{\sin x} = \cot x$.
So, the LHS simplifies to:

\tan(\theta + \alpha)\cot\alpha

Conclude the proof. From step 2, we had:

\frac{\sin(\theta + 2\alpha) + \sin\theta}{\sin(\theta + 2\alpha) - \sin\theta} = \frac{m + n}{m - n}

And from step 4, we found that the left-hand side simplifies to $\tan(\theta + \alpha)\cot\alpha$.
Therefore, we have:

\tan(\theta + \alpha)\cot\alpha = \frac{m + n}{m - n}

This completes the proof.

✓Final answer

m\sin\theta = n\sin(\theta + 2\alpha)

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Q.If $m\sin\theta = n\sin(\theta + 2\alpha)$ , then prove that $\tan(\theta + \alpha)\cot\alpha = \dfrac{m + n}{m - n}$ .

Unlock everything free for 14 days