If $\alpha + \beta = \dfrac{\pi}{4}$, then the value of $(1 + \tan\alpha)(1 + \tan\beta)$ is (A) $1$ (B) $2$ (C) $-2$ (D) Not defined

If α + β = (π)/(4), then the value of (1 + tanα)(1 + tanβ) is (A) 1 (B) 2 (C) -2 (D) Not defined

NCERT Exemplar · Q51

\alpha + \beta = \dfrac{\pi}{4}

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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 applying the tangent addition formula to the given condition $\alpha + \beta = \frac{\pi}{4}$ , we derive a relationship between $\tan\alpha$ and $\tan\beta$ that directly simplifies the expression $(1 + \tan\alpha)(1 + \tan\beta)$ to 2.

The core idea here is to leverage the given sum of angles, $\alpha + \beta = \frac{\pi}{4}$ , to find a relationship between $\tan\alpha$ and $\tan\beta$ . When you encounter a problem involving a sum of angles and tangent functions, the tangent addition formula should immediately come to mind.

The expression we need to evaluate, $(1 + \tan\alpha)(1 + \tan\beta)$ , looks like it might simplify if we can find a way to relate the sum $\tan\alpha + \tan\beta$ and the product $\tan\alpha \tan\beta$ . The tangent addition formula is perfectly suited for this, as it involves exactly these terms.

By taking the tangent of both sides of the given condition, we can substitute the known value of $\tan(\frac{\pi}{4})$ and then rearrange the formula to get a direct link between the sum and product of $\tan\alpha$ and $\tan\beta$ . Once we have this link, expanding the target expression $(1 + \tan\alpha)(1 + \tan\beta)$ will reveal that it can be directly simplified using the relationship we derived. This type of problem is a classic application of trigonometric identities, designed to test your ability to recognize and apply the correct formula to simplify an expression.

Here is the step-by-step solution:

Start with the given condition. We are given that the sum of angles $\alpha$ and $\beta$ is $\frac{\pi}{4}$ radians.

\alpha + \beta = \frac{\pi}{4}

Apply the tangent function to both sides. To introduce $\tan\alpha$ and $\tan\beta$ into the equation, we take the tangent of both sides of the given condition:

\tan(\alpha + \beta) = \tan\left(\frac{\pi}{4}\right)

Use the tangent addition formula. Recall the tangent addition formula: $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$

\frac{\tan\alpha + \tan\beta}{1 - \tan\alpha \tan\beta} = \tan\left(\frac{\pi}{4}\right)

Substitute the known value of $\tan(\frac{\pi}{4})$ . We know that $\tan\left(\frac{\pi}{4}\right) = 1$ . Substituting this value into the equation:

\frac{\tan\alpha + \tan\beta}{1 - \tan\alpha \tan\beta} = 1

Rearrange the equation to find a relationship. Multiply both sides by $(1 - \tan\alpha \tan\beta)$ :

\tan\alpha + \tan\beta = 1 - \tan\alpha \tan\beta

Now, move the term $\tan\alpha \tan\beta$ from the right side to the left side:

\tan\alpha + \tan\beta + \tan\alpha \tan\beta = 1

This equation provides a crucial relationship between the sum and product of $\tan\alpha$ and $\tan\beta$. …

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Q.If α+β=π4\alpha + \beta = \dfrac{\pi}{4}α+β=4π​, then the value of (1+tan⁡α)(1+tan⁡β)(1 + \tan\alpha)(1 + \tan\beta)(1+tanα)(1+tanβ) is (A) 111 (B) 222 (C) −2-2−2 (D) Not defined

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Q.If $\alpha + \beta = \dfrac{\pi}{4}$ , then the value of $(1 + \tan\alpha)(1 + \tan\beta)$ is
(A) $1$
(B) $2$
(C) $-2$
(D) Not defined

Unlock everything free for 14 days