Question

If A, B, C, D are four angles then sin (A-B).cos (A+ B) +sin (B-C). cos (B+C) +sin(C-D). cos (C+D) +sin(D-A). cos (D+A) =
$
  A)0 \\
  B)1 \\
  C)4\sin A\sin B\sin C\sin D \\
  D)4\cos A\cos B\cos C\cos D \\
$

Answer 1

Hint: In order to solve this type of question, we will firstly multiply and divide the whole expression by 2. As the expression is too long, we will solve it into parts by using sum angles formula i.e. $2\sin (X)\cos (Y) = \sin (X + Y) + \sin (X - Y)$. Then we will add it and simplify to get the desired result.

Complete step by step answer:
Given ,
A, B, C, D are the four angles.
$\sin (A - B)\cos (A + B) + \sin (B - C)\cos (B + C) + \sin (C - D)\cos (C + D) + \sin (D - A)\cos (D + A)$
Firstly, we will multiply and divide whole equation by 2 , as we can form the angle sum formulas from them, we get
As the equation is too long, let us multiply the whole equation in four parts and then add the parts as given in the question.
$\dfrac{2}{2} \times \left[ {\sin (A - B)\cos (A + B) + \sin (B - C)\cos (B + C) + \sin (C - D)\cos (C + D) + \sin (D - A)\cos (D + A)} \right]$
First part:- $\sin (A - B)\cos (A + B)$
Here when we will multiply the equation with 2, We get,
$2\sin (A - B)\cos (A + B)$
As we know that $2\sin (X)\cos (Y) = \sin (X + Y) + \sin (X - Y)$ , we get
$
  \sin (A - B + A + B) + \sin (A - B - A - B) \\
   = \sin (2A) + \sin ( - 2B) \\
   = \sin (2A) - \sin (2B) \\
$

Second part:- $\sin (B - C)\cos (B + C)$
Here when we will multiply the equation with 2, We get,
$2\sin (B - C)\cos (C + B)$
As we know that $2\sin (X)\cos (Y) = \sin (X + Y) + \sin (X - Y)$ , we get
$
  \sin (B - C + B + C) + \sin (B - C - B - C) \\
   = \sin (2B) + \sin ( - 2C) \\
   = \sin (2B) - \sin (2C) \\
$

Third part:- $\sin (C - D)\cos (C + D)$
Here when we will multiply the equation with 2, We get,
$2\sin (C - D)\cos (C + D)$
As we know that $2\sin (X)\cos (Y) = \sin (X + Y) + \sin (X - Y)$ , we get
$
  \sin (C - D + C + D) + \sin (C - D - C - D) \\
   = \sin (2C) + \sin ( - 2D) \\
   = \sin (2C) - \sin (2D) \\
$

Fourth part:- $\sin (D - A)\cos (D + A)$
Here when we will multiply the equation with 2, We get,
$2\sin (D - A)\cos (D + A)$
As we know that $2\sin (X)\cos (Y) = \sin (X + Y) + \sin (X - Y)$ , we get
$
  \sin (D - A + D + A) + \sin (D - A - D - A) \\
   = \sin (2D) + \sin ( - 2A) \\
   = \sin (2D) - \sin (2A) \\
$

Now we will add all the four parts,
$
  \dfrac{1}{2}\left[ {\sin 2A - \sin 2B + \sin 2B - \sin 2C + \sin 2C - \sin 2D + \sin 2D - \sin 2A} \right] \\
   = \dfrac{1}{2}\left[ 0 \right] \\
   = 0 \\
$
As all the terms got cancelled out , we get the final answer as 0

So, the correct answer is “Option A”.

Note: While solving this question, we must concentrate on why we divided the equation with the same number and multiply it also with the same number. We did this because if we will multiply and divide the equation with the same number, we can create a condition for sum angle identity . Also we must know the common trigonometry formulas especially the identities.