Question

: Let $ \tan x = m\tan y $ , then $ \sin \left( {x + y} \right) $ equal:
A. $ \left( {\dfrac{{m + 1}}{{m - 1}}} \right)\sin \left( {x - y} \right) $
B. $ \left( {\dfrac{{m - 1}}{{m + 1}}} \right)\sin \left( {x - y} \right) $
C. $ \sqrt {1 + {m^2}} \sin \left( {x - y} \right) $
D. $ \dfrac{{2m + 1}}{{2m - 1}}\sin \left( {x - y} \right) $

Answer 1

Hint: Simplify the term $ \sin \left( {x + y} \right) $ using the formula for sum of angles for sine function and then check the value of $ \sin \left( {x - y} \right) $ and compare them both and get the value of $ \sin \left( {x + y} \right) $ in terms of $ \sin \left( {x - y} \right) $ .

Complete step-by-step answer:
As per statement it is given that $ \tan x = m\tan y $ .
Use the trigonometric formula for the sum of angles for the trigonometric function sine given as $ \sin \left( {x + y} \right) = \sin x\cos y + \cos x\sin y $ .
Simplify the term $ \sin \left( {x + y} \right) $ with the help of the trigonometric formula mentioned above.
 $ \sin \left( {x + y} \right) = \sin x\cos y + \cos x\sin y $
Divide the right hand side of the equation by $ \cos x\cos y $ and also multiply right hand side by the same term.
 $
  \sin \left( {x + y} \right) = \cos x\cos y\left( {\dfrac{{\sin x\cos y}}{{\cos x\cos y}} + \dfrac{{\cos x\sin y}}{{\cos x\cos y}}} \right) \\
   = \cos x\cos y\left( {\tan x + \tan y} \right) \;
  $
As given $ \tan x = m\tan y $ substitute $ m\tan y $ for the value of $ \tan x $ in the above equation.
 $
  \sin \left( {x + y} \right) = \cos x\cos y\left( {\tan x + \tan y} \right) \\
   = \cos x\cos y\left( {m\tan y + \tan y} \right) \\
   = \cos x\cos y\tan y\left( {m + 1} \right) \\
   = \cos x\sin y\left( {m + 1} \right)\;\;\; \ldots \left( 1 \right) \;
  $
Now simplify the term $ \sin \left( {x - y} \right) $ by the trigonometric formula.
 $ \sin \left( {x - y} \right) = \sin x\cos y - \cos x\sin y $
Divide the right hand side of the equation by $ \cos x\cos y $ and also multiply right hand side by the same term.
 $
  \sin \left( {x - y} \right) = \cos x\cos y\left( {\dfrac{{\sin x\cos y}}{{\cos x\cos y}} - \dfrac{{\cos x\sin y}}{{\cos x\cos y}}} \right) \\
   = \cos x\cos y\left( {\tan x - \tan y} \right) \\
   = \cos x\cos y\left( {m\tan y - \tan y} \right) \\
   = \cos x\sin y\left( {m - 1} \right) \;
  $
As $ \sin \left( {x - y} \right) = \cos x\sin y\left( {m - 1} \right) $ . So, we can say that $ \cos x\sin y = \dfrac{{\sin \left( {x - y} \right)}}{{m - 1}} $ .
Substitute the value of $ \cos x\sin y $ in the equation $ \left( 1 \right) $ .
 $
  \sin \left( {x + y} \right) = \cos x\sin y\left( {m + 1} \right) \\
   = \dfrac{{\sin \left( {x - y} \right)}}{{\left( {m - 1} \right)}}\left( {m + 1} \right) \\
   = \left( {\dfrac{{m + 1}}{{m - 1}}} \right)\sin \left( {x - y} \right) \;
  $
So, the value of $ \sin \left( {x + y} \right) $ is equal to $ \left( {\dfrac{{m + 1}}{{m - 1}}} \right)\sin \left( {x - y} \right) $ .
So, the correct answer is “Option A”.

Note: Use the trigonometric formulas for the sum and difference of the angles for the trigonometric function sine as $ \sin \left( {x + y} \right) = \sin x\cos y + \cos x\sin y $ and $ \sin \left( {x - y} \right) = \sin x\cos y - \cos x\sin y $ . Compare both the values as there is a common factor in both the simplifications.