Question

Prove the equation given below:
$\dfrac{{\cos 7x + cos5x}}{{\sin 7x - \sin 5x}} = \cot x$

Answer 1

Hint-We will use trigonometric identities to solve such types of questions. In this problem we will use an identity which will convert addition operation into multiplication.
$
  1)\cos c + \cos d = 2\cos (\dfrac{{c + d}}{2})\cos (\dfrac{{c - d}}{2}) \\
  2)\sin c - \sin d = 2\cos (\dfrac{{c + d}}{2})\sin (\dfrac{{c - d}}{2}) \\
 $

Complete step-by-step answer:


Given equation:
$ \Rightarrow \dfrac{{\cos 7x + cos5x}}{{\sin 7x - \sin 5x}} = \cot x$
Taking L.H.S to solve the problem
$ = \dfrac{{\cos 7x + cos5x}}{{\sin 7x - \sin 5x}}$
As we know that
$
  \cos c + \cos d = 2\cos (\dfrac{{c + d}}{2})\cos (\dfrac{{c - d}}{2}) \\
  \sin c - \sin d = 2\cos (\dfrac{{c + d}}{2})\sin (\dfrac{{c - d}}{2}) \\
 $
By applying these identities in the given equation and simplifying it further, we obtain
$
   = \dfrac{{2\cos (\dfrac{{7x + 5x}}{2})\cos (\dfrac{{7x - 5x}}{2})}}{{2\cos (\dfrac{{7x + 5x}}{2})\sin (\dfrac{{7x - 5x}}{2})}} \\
   = \dfrac{{2\cos (6x)\cos (x)}}{{2\cos (6x)\sin (x)}} \\
   = \dfrac{{\cos x}}{{\sin x}}{\text{ }}\left[ {{\text{As we know }}\dfrac{{\cos x}}{{\sin x}} = \cot x{\text{ }}} \right] \\
   = \cot x \\
 $
Hence, finally we get L.H.S = $\cot x$ which is equal to R.H.S. So the equation is proved.

Note- To solve these types of questions, you must remember all trigonometric expressions and should have a good grasp on algebra. In this question we converted the sum into multiplication using the sum of products formula. Similarly in other questions we may have to convert the product into sum, so these formulas will help to convert them.