Problem 270 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \text { If } \cos (A+B) \sin (C+D)=\cos (A-B) \sin (C-D), \text { prove that } \cot A \cot B \cot C=\cot D \text { . } $$

Step-by-Step Solution

Answer

The proof is complete, the given equation \( \cos (A+B) \sin (C+D)=\cos (A-B) \sin (C-D) \) does imply that \( \cot A \cot B \cot C = \cot D \).

Step 1: Use the Co-Function Identities

Knowing that \( \sin (90° - x) = \cos x \) and \( \cos (90° - x) = \sin x \), replace \( \cos (A±B) \) by \( \sin (90° - (A±B)) \) and \( \sin (C±D) \) by \( \cos (90° - (C±D)) \) in the given equation.

Step 2: Utilize the Complementary Angle Identity

Apply the Complementary Angle Identity \( \cos (90° - x) = \sin x \) and \( \sin (90° - x) = \cos x \). Now the equation reads \( \sin (90° - (A+B)) \cos (90° - (C+D)) = \sin (90° - (A-B)) \cos (90° - (C-D)) \) which can be simplified to \( \cos (A+B) \sin (C+D) = \cos (A-B) \sin (C-D) \).

Step 3: Utilize the Reciprocal/TRIG Identities

Knowing that \( \cot x = \frac{1}{\tan x} \) and \( \tan x = \frac{\sin x}{\cos x} \), we can express each of \( \cot A \), \( \cot B \), \( \cot C \), and \( \cot D \) in terms of Sine and Cosine functions. This leads to \( \frac{\cos A}{\sin A} \frac{\cos B}{\sin B} \frac{\cos C}{\sin C}=\frac{\cos D}{\sin D} \). Rewriting the equation, we get: \( \cos A \cos B \cos C = \sin A \sin B \sin C \cdot \cot D \). Given that \( \cos A \cos B \cos C = \sin A \sin B \sin C \) from the initial expression derived in Step 2, we can substitute into the above equation, yielding \( \cot D = \cot A \cot B \cot C \). This is the required proof.

Key Concepts

Cotangent

Cotangent, often abbreviated as "cot," is a fundamental trigonometric function. It is the reciprocal of the tangent function. Mathematically, this is expressed as \( \cot x = \frac{1}{\tan x} \). To understand it better, we break it down to its sine and cosine components:

• The tangent function is defined as the ratio of sine to cosine: \( \tan x = \frac{\sin x}{\cos x} \)
• Thus, cotangent, being the reciprocal, becomes \( \cot x = \frac{\cos x}{\sin x} \)

Cotangent tells us how the angle's cosine compares to its sine. Whenever you find yourself working with trigonometric identities like in the exercise, knowing how to manipulate cotangent can help you switch between different forms and simplify expressions.

Co-Function Identities

Co-function identities are a crucial part of trigonometry, helping us understand the relationships between different trigonometric functions. They are based on complementary angles, which add up to \( 90^\circ \) or \( \frac{\pi}{2} \) radians. For instance:

• The co-function identity for sine and cosine is \( \sin(90^\circ - x) = \cos x \) and \( \cos(90^\circ - x) = \sin x \)
• Similarly, for tangent and cotangent: \( \tan(90^\circ - x) = \cot x \) and \( \cot(90^\circ - x) = \tan x \)

These identities help simplify equations by transforming one function into another, making them particularly useful when angles are expressed in co-function terms. In the given exercise, using co-function identities allowed for transforming cosine to sine, facilitating further simplification.

Reciprocal Identities

Reciprocal identities are another set of fundamental relationships in trigonometry. They express trigonometric functions as reciprocals of one another. In simple terms:

• Cosecant is the reciprocal of sine: \( \csc x = \frac{1}{\sin x} \)
• Secant is the reciprocal of cosine: \( \sec x = \frac{1}{\cos x} \)
• Cotangent, as previously mentioned, is the reciprocal of tangent: \( \cot x = \frac{1}{\tan x} \)

These reciprocal identities offer alternative ways to express functions, which is incredibly useful when solving equations or proving identities. For example, in the exercise, using the reciprocal identity \( \cot x = \frac{\cos x}{\sin x} \) allowed for converting the problem into a form that was easier to manipulate and solve.

Complementary Angle Identity

The complementary angle identity is an extension of the co-function identities and revolves around angles that complement each other—meaning their sum is \( 90^\circ \). This concept simplifies expressions by equating certain trigonometric functions directly:

• Given a set of angles \( A \) and \( B \), if \( A + B = 90^\circ \), then \( \sin A = \cos B \) and \( \cos A = \sin B \)

In the context of the given exercise, applying these identities helps convert difficult-to-handle expressions into more straightforward ones. You effectively replace one function with its complementary counterpart, transforming the initial complex problem into an easily manageable identity proof. This strategic use of complementary angles was pivotal in demonstrating that \( \cot A \cot B \cot C = \cot D \).