If $\cos(\theta + \phi) = m\cos(\theta - \phi)$, then prove that $\tan\theta = \dfrac{1 - m}{1 + m}\cot\phi$.

If cos(θ + φ) = mcos(θ - φ), then prove that tanθ – Class 11 Mathematics

NCERT Exemplar · Q21

\cos(\theta + \phi) = m\cos(\theta - \phi)

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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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By rearranging the given equation into a ratio and applying the Componendo and Dividendo rule along with sum-to-product formulas, we can directly derive the required identity. The final result is $\tan\theta = \dfrac{1 - m}{1 + m}\cot\phi$ .

The problem asks us to prove a trigonometric identity starting from a given equation. The initial equation, $\cos(\theta + \phi) = m\cos(\theta - \phi)$ , involves cosines of sums and differences of angles. The target identity, $\tan\theta = \dfrac{1 - m}{1 + m}\cot\phi$ , involves tangents and cotangents. This suggests that we need to transform the cosine terms into tangent/cotangent forms.

A common and efficient strategy when dealing with an equation of the form $A = mB$ (which can be written as $\frac{A}{B} = \frac{m}{1}$ ) is to use the Componendo and Dividendo rule. This rule allows us to introduce sums and differences of the numerator and denominator, which, in trigonometry, often pairs perfectly with sum-to-product or product-to-sum formulas. In this specific case, $\cos(X) \pm \cos(Y)$ expressions will naturally arise, which can then be simplified into products of sines and cosines, leading directly to tangent or cotangent ratios.

Let's work through the proof step-by-step.

Rearrange the given equation into a ratio. We are given the equation:

\cos(\theta + \phi) = m\cos(\theta - \phi)

To prepare for Componendo and Dividendo, we can express this as a ratio:

\frac{\cos(\theta + \phi)}{\cos(\theta - \phi)} = \frac{m}{1}

Apply the Componendo and Dividendo rule. $\frac{a}{b} = \frac{c}{d}$

\frac{\cos(\theta + \phi) + \cos(\theta - \phi)}{\cos(\theta + \phi) - \cos(\theta - \phi)} = \frac{m + 1}{m - 1}

> [!TIP]
> Componendo and Dividendo is a powerful algebraic tool. It's particularly effective in trigonometry when you have ratios of sums/differences of angles, as it often sets up expressions that can be simplified using sum-to-product or product-to-sum identities.

3. Use sum-to-product trigonometric identities.

We need to simplify the numerator and denominator of the left-hand side using the sum-to-product formulas for cosine:

> [!FORMULA]

> $\cos A + \cos B = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$

> $\cos A - \cos B = -2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$

Let $A = \theta + \phi$ and $B = \theta - \phi$ .

Then, we calculate the sum and difference of these angles:

$A+B = (\theta + \phi) + (\theta - \phi) = 2\theta$

$A-B = (\theta + \phi) - (\theta - \phi) = 2\phi$

So, $\frac{A+B}{2} = \theta$ and $\frac{A-B}{2} = \phi$ .

Now, substitute these into the sum-to-product formulas:
*   Numerator: $\cos(\theta + \phi) + \cos(\theta - \phi) = 2\cos\theta \cos\phi$ …

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Q.If cos⁡(θ+ϕ)=mcos⁡(θ−ϕ)\cos(\theta + \phi) = m\cos(\theta - \phi)cos(θ+ϕ)=mcos(θ−ϕ), then prove that tan⁡θ=1−m1+mcot⁡ϕ\tan\theta = \dfrac{1 - m}{1 + m}\cot\phitanθ=1+m1−m​cotϕ.

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Q.If $\cos(\theta + \phi) = m\cos(\theta - \phi)$ , then prove that $\tan\theta = \dfrac{1 - m}{1 + m}\cot\phi$ .

Unlock everything free for 14 days