Problem 33 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

If \(\left(1+\tan 1^{\circ}\right)\left(1+\tan 2^{\circ}\right) \ldots\left(1+\tan 45^{\circ}\right)=2^{n}\) then the value of \(n\) is (a) 20 (b) 21 (c) 22 (d) 23

Step-by-Step Solution

Answer

(c) 22

Step 1: Identify complementary angle pairs

A trick that trigonometric problems involving sum and product often use is to consider complementary angles. Recall that for any acute angle \(\theta\), we have \(\tan (45 - \theta) = \tan (45 + \theta)\). Therefore, pair up terms in the given expression whose angles sum up to 45 degrees, specifically \(1 + \tan(1^{\circ}), 1 + \tan(2^{\circ}), \ldots, 1 + \tan(22^{\circ})\) and their corresponding complementary terms \(1 + \tan(44^{\circ}), 1 + \tan(43^{\circ}), \ldots, 1 + \tan(23^{\circ})\), in addition to the central term \(1 + \tan(45^{\circ})\).

Step 2: Simplify each pair

Upon simplification, each pair \((1 + \tan a)(1 + \tan b)\) equals \(2\) where \(a + b = 45^{\circ}\). This is because \(\tan a = \tan (45 - (45 - a)) = \tan (45 - b)\), the tangent of a complement of an angle equals to the tangent of the angle. Hence, \(1 + \tan a = 1 + \tan (45 - b) = (1 + \tan b)/ \tan b = 2\).

Step 3: Calculate the total product

Putting these all together, the total product \((1 + \tan 1^{\circ})(1 + \tan 2^{\circ}) \ldots (1 + \tan 45^{\circ})\) equals \(2^{22}\) as there are 22 multiplication pairs of terms, each equals to \(2\).

Key Concepts

Complementary Angles

Complementary angles are two angles whose measures add up to 90 degrees. This concept is pivotal in trigonometry, especially when dealing with problems like the one given in our exercise. In trigonometric identities, complementary angles often reveal interesting relationships between the sine, cosine, and tangent functions.
For instance:

• \( \sin(90^{\circ} - \theta) = \cos(\theta) \)
• \( \cos(90^{\circ} - \theta) = \sin(\theta) \)
• \( \tan(90^{\circ} - \theta) = \cot(\theta) \)

In the original exercise, we leverage the identity of \( \tan(\theta) \) and its complement to simplify the product of tangents. By recognizing that the angles are complementary, such as \( 1^{\circ} \) and \( 44^{\circ} \) which sum up to 45 degrees, we use the relationship: \( \tan(45^{\circ} - \theta) = \tan \theta \) to tackle the expression effectively.

Product of Tangents

The product of tangents is an intriguing part of trigonometry, and understanding it fully can unveil elegant solutions to complex problems. When we talk about the product of tangents in the context of trigonometric equations, we're generally referring to multiplying multiple tangent terms together.
In the given exercise, we are tasked with finding the product:

• \((1 + \tan 1^{\circ})(1 + \tan 2^{\circ}) \ldots (1 + \tan 45^{\circ})\)

The key realization here is that each pair of terms in this product simplifies to 2 when you pair complementary angles, such as \( (1+ \tan 1^{\circ})(1 + \tan 44^{\circ}) = 2 \). By simplifying each pair systematically, the calculation becomes straightforward.
This simplification leads to a product that can be represented as a power of 2 and beautifully shows how trigonometric identities help in solving such compounded expressions.

Trigonometric Equations

Trigonometric equations often involve finding unknown angles or values by employing various trigonometric identities. Solving these equations requires familiarity with identities and the relationships between trigonometric functions. For instance, understanding how complementary angles interact can reduce seemingly complex equations to simple expressions.
In our original exercise, the trigonometric equation involves a product expression set equal to a power of 2:

• \((1+ \tan 1^{\circ})(1+\tan 2^{\circ}) \cdots (1+\tan 45^{\circ}) = 2^n\)

By breaking down the equation into simpler parts using complementary angle pairs, we solve for \(n\) as each product pair results in 2. This approach demonstrates how systematically replacing and simplifying terms within trigonometric equations assists in finding solutions efficiently.
Mastering these methods equips students with powerful tools to confront more challenging trigonometric problems in their studies.