Solveeit Logo
Login

Question

Question: : Let \( \tan x = m\tan y \) , then \( \sin \left( {x + y} \right) \) equal: A. \( \left( {\dfrac{...

: Let tan⁡x=mtan⁡y\tan x = m\tan y , then sin⁡(x+y)\sin \left( {x + y} \right) equal:
A. (m+1m−1)sin⁡(x−y)\left( {\dfrac{{m + 1}}{{m - 1}}} \right)\sin \left( {x - y} \right)
B. (m−1m+1)sin⁡(x−y)\left( {\dfrac{{m - 1}}{{m + 1}}} \right)\sin \left( {x - y} \right)
C. 1+m2sin⁡(x−y)\sqrt {1 + {m^2}} \sin \left( {x - y} \right)
D. 2m+12m−1sin⁡(x−y)\dfrac{{2m + 1}}{{2m - 1}}\sin \left( {x - y} \right)

Explanation

Solution

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

Complete step-by-step answer :
As per statement it is given that tan⁡x=mtan⁡y\tan x = m\tan y .
Use the trigonometric formula for the sum of angles for the trigonometric function sine given as sin⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡y\sin \left( {x + y} \right) = \sin x\cos y + \cos x\sin y .
Simplify the term sin⁡(x+y)\sin \left( {x + y} \right) with the help of the trigonometric formula mentioned above.
sin⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡y\sin \left( {x + y} \right) = \sin x\cos y + \cos x\sin y
Divide the right hand side of the equation by cos⁡xcos⁡y\cos x\cos y and also multiply right hand side by the same term.
sin⁡(x+y)=cos⁡xcos⁡y(sin⁡xcos⁡ycos⁡xcos⁡y+cos⁡xsin⁡ycos⁡xcos⁡y) =cos⁡xcos⁡y(tan⁡x+tan⁡y)   \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=mtan⁡y\tan x = m\tan y substitute mtan⁡ym\tan y for the value of tan⁡x\tan x in the above equation.
sin⁡(x+y)=cos⁡xcos⁡y(tan⁡x+tan⁡y) =cos⁡xcos⁡y(mtan⁡y+tan⁡y) =cos⁡xcos⁡ytan⁡y(m+1) =cos⁡xsin⁡y(m+1)      …(1)   \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⁡(x−y)\sin \left( {x - y} \right) by the trigonometric formula.
sin⁡(x−y)=sin⁡xcos⁡y−cos⁡xsin⁡y\sin \left( {x - y} \right) = \sin x\cos y - \cos x\sin y
Divide the right hand side of the equation by cos⁡xcos⁡y\cos x\cos y and also multiply right hand side by the same term.
sin⁡(x−y)=cos⁡xcos⁡y(sin⁡xcos⁡ycos⁡xcos⁡y−cos⁡xsin⁡ycos⁡xcos⁡y) =cos⁡xcos⁡y(tan⁡x−tan⁡y) =cos⁡xcos⁡y(mtan⁡y−tan⁡y) =cos⁡xsin⁡y(m−1)   \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⁡(x−y)=cos⁡xsin⁡y(m−1)\sin \left( {x - y} \right) = \cos x\sin y\left( {m - 1} \right) . So, we can say that cos⁡xsin⁡y=sin⁡(x−y)m−1\cos x\sin y = \dfrac{{\sin \left( {x - y} \right)}}{{m - 1}} .
Substitute the value of cos⁡xsin⁡y\cos x\sin y in the equation (1)\left( 1 \right) .
sin⁡(x+y)=cos⁡xsin⁡y(m+1) =sin⁡(x−y)(m−1)(m+1) =(m+1m−1)sin⁡(x−y)   \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⁡(x+y)\sin \left( {x + y} \right) is equal to (m+1m−1)sin⁡(x−y)\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⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡y\sin \left( {x + y} \right) = \sin x\cos y + \cos x\sin y and sin⁡(x−y)=sin⁡xcos⁡y−cos⁡xsin⁡y\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.