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Question: Prove that \[\dfrac{{\tan x + \tan y}}{{\tan x - \tan y}} = \dfrac{{\sin (x + y)}}{{\sin (x - y)}}\]...

Prove that tan⁡x+tan⁡ytan⁡x−tan⁡y=sin⁡(x+y)sin⁡(x−y)\dfrac{{\tan x + \tan y}}{{\tan x - \tan y}} = \dfrac{{\sin (x + y)}}{{\sin (x - y)}}

Explanation

Solution

To solve this problem, we use the concepts of trigonometry and their properties. We will also use some trigonometric identities like sin⁡(A±B)=sin⁡Acos⁡B±sin⁡Bcos⁡A\sin (A \pm B) = \sin A\cos B \pm \sin B\cos A and also some basic formula like tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}. We will prove this equation by equating the LHS and the RHS.

Complete step by step answer:
Here, to solve this problem, we will consider the left-hand side of the given equation and simplify that using some trigonometric identities and check whether we are getting the same expression present on the right-hand side or not.
So, simply, we have to prove the Left Hand Side is equal to the Right Hand Side i.e., LHS=RHSLHS = RHS.
Now take the left-hand side.
LHS⇒tan⁡x+tan⁡ytan⁡x−tan⁡yLHS \Rightarrow \dfrac{{\tan x + \tan y}}{{\tan x - \tan y}}
We know that tangent value of an angle is the ratio of sine of that angle to cosine of that angle, i.e., tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}
LHS⇒sin⁡xcos⁡x+sin⁡ycos⁡ysin⁡xcos⁡x−sin⁡ycos⁡yLHS \Rightarrow \dfrac{{\dfrac{{\sin x}}{{\cos x}} + \dfrac{{\sin y}}{{\cos y}}}}{{\dfrac{{\sin x}}{{\cos x}} - \dfrac{{\sin y}}{{\cos y}}}}
Now, when we simplify these fractions by taking LCM, we get,
⇒sin⁡xcos⁡y+sin⁡ycos⁡xcos⁡xcos⁡ysin⁡xcos⁡y−sin⁡ycos⁡xcos⁡xcos⁡y\Rightarrow \dfrac{{\dfrac{{\sin x\cos y + \sin y\cos x}}{{\cos x\cos y}}}}{{\dfrac{{\sin x\cos y - \sin y\cos x}}{{\cos x\cos y}}}}
There are fractions in both numerator and denominator, and the denominators of both the fractions are the same. So, they get cancelled.
⇒sin⁡xcos⁡y+sin⁡ycos⁡xsin⁡xcos⁡y−sin⁡ycos⁡x\Rightarrow \dfrac{{\sin x\cos y + \sin y\cos x}}{{\sin x\cos y - \sin y\cos x}}
Now, we know the formulas
sin⁡(A+B)=sin⁡Acos⁡B+sin⁡Bcos⁡A\sin (A + B) = \sin A\cos B + \sin B\cos A and
sin⁡(A−B)=sin⁡Acos⁡B−sin⁡Bcos⁡A\sin (A - B) = \sin A\cos B - \sin B\cos A
These are trigonometric identities related to sum or difference of angles.
So, from these two formulas, we can conclude that,
⇒sin⁡(x+y)sin⁡(x−y)=RHS\Rightarrow \dfrac{{\sin (x + y)}}{{\sin (x - y)}} = RHS
So, we concluded that, LHS=RHSLHS = RHS
Hence, we proved the equation.

Note: We should note that this type of problem can be solved using simplification of RHS as well that won’t affect the solution.
There are two variables present in the left side expression, so we get the resulting expression also in two variables.
Substitute some values like x=π3x = \dfrac{\pi }{3} and y=π6y = \dfrac{\pi }{6} in the expression on LHS, and check whether you get the same value, when you substitute these values in RHS or not. This is also a verification method.