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Question: How do you prove \[\left[ {\dfrac{{\sin (x + y)}}{{\sin (x - y)}}} \right] = \left[ {\dfrac{{\tan (x...

How do you prove [sin⁡(x+y)sin⁡(x−y)]=[tan⁡(x)+tan⁡(y)tan⁡(x)−tan⁡(y)]\left[ {\dfrac{{\sin (x + y)}}{{\sin (x - y)}}} \right] = \left[ {\dfrac{{\tan (x) + \tan (y)}}{{\tan (x) - \tan (y)}}} \right]?

Explanation

Solution

According to the question, we will first take out the LHS part and RHS part. Then, we will try to solve one of the parts. We can start solving either the LHS part or RHS part. Solving and simplifying one of the parts will end up giving us the second part.

Complete step-by-step solution:
First, we will take out the LHS part and RHS part from the given question, and we will get:
LHS=[sin⁡(x+y)sin⁡(x−y)] = \left[ {\dfrac{{\sin (x + y)}}{{\sin (x - y)}}} \right]
RHS=[tan⁡(x)+tan⁡(y)tan⁡(x)−tan⁡(y)] = \left[ {\dfrac{{\tan (x) + \tan (y)}}{{\tan (x) - \tan (y)}}} \right]
Now, we will try to solve the RHS part first, and we will get:
RHS=[tan⁡(x)+tan⁡(y)tan⁡(x)−tan⁡(y)] = \left[ {\dfrac{{\tan (x) + \tan (y)}}{{\tan (x) - \tan (y)}}} \right]
We can rewrite this part as:
RHS=[tan⁡x+tan⁡ytan⁡x−tan⁡y] = \left[ {\dfrac{{\tan x + \tan y}}{{\tan x - \tan y}}} \right]
We will try to expand the RHS part. We will try to expand tan⁡x \tan x\,and tan⁡y\tan y from the basic trigonometric formula that says:
tan⁡a=sin⁡acos⁡a\tan a = \dfrac{{\sin a}}{{\cos a}}
tan⁡x+tan⁡y=sin⁡xcos⁡x+sin⁡ycos⁡y\tan x + \tan y = \dfrac{{\sin x}}{{\cos x}} + \dfrac{{\sin y}}{{\cos y}}
Now, we will simplify this part, and we will get:
⇒tan⁡x+tan⁡y=sin⁡xcos⁡y+cos⁡xsin⁡ycos⁡xcos⁡y\Rightarrow \tan x + \tan y = \dfrac{{\sin x\cos y + \cos x\sin y}}{{\cos x\cos y}}
We know that sin⁡xcos⁡y+cos⁡xsin⁡y=sin⁡(x+y)\sin x\cos y + \cos x\sin y = \sin (x + y). So, we will put this value in our equation, and we will get:
⇒tan⁡x+tan⁡y=sin⁡(x+y)cos⁡xcos⁡y\Rightarrow \tan x + \tan y = \dfrac{{\sin (x + y)}}{{\cos x\cos y}}
Similarly, when we solve tan⁡x−tan⁡y\tan x - \tan y, we get:
⇒tan⁡x−tan⁡y=sin⁡(x−y)cos⁡xcos⁡y\Rightarrow \tan x - \tan y = \dfrac{{\sin (x - y)}}{{\cos x\cos y}}
Now, we got two equations of tan⁡x+tan⁡y\tan x + \tan yand tan⁡x−tan⁡y\tan x - \tan y. We will try to divide both the equations, and we will get:
⇒tan⁡x+tan⁡ytan⁡x−tan⁡y=sin⁡(x+y)cos⁡xcos⁡ysin⁡(x−y)cos⁡xcos⁡y\Rightarrow \dfrac{{\tan x + \tan y}}{{\tan x - \tan y}} = \dfrac{{\dfrac{{\sin (x + y)}}{{\cos x\cos y}}}}{{\dfrac{{\sin (x - y)}}{{\cos x\cos y}}}}
Here, we will cancel the like terms, and try to simplify the equation, and we will get:
⇒tan⁡x+tan⁡ytan⁡x−tan⁡y=sin⁡(x+y)sin⁡(x−y)\Rightarrow \dfrac{{\tan x + \tan y}}{{\tan x - \tan y}} = \dfrac{{\sin (x + y)}}{{\sin (x - y)}}
So, here we get our LHS part. Our LHS part is sin⁡(x+y)sin⁡(x−y)\dfrac{{\sin (x + y)}}{{\sin (x - y)}}.

Hence, we have proved that L.H.S = R.H.S

Note: In this problem the main part is to remember the formula of tanxtan x i.e, tanx=sinxcosxtan x =\dfrac{sin x}{cos x} because the whole solution is to simplify the given term so here the given equation has tan x terms so to simplify the whole equation we need tanxtan x formula and here we have also used one more formula which simplify it more i.e, sin⁡xcos⁡y+cos⁡xsin⁡y=sin⁡(x+y)\sin x\cos y + \cos x\sin y = \sin (x + y). The above method was easy, and the question gets solved quickly. But there is another method also to solve this question. Instead of first solving the RHS part, we can start solving the LHS part as well. But here we should prefer solving the RHS part only, because it is easy to expand the RHS part.