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

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

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)cos⁡(x+y)+cos⁡(x−y) = \dfrac{{\sin (x + y) - \sin (x - y)}}{{\cos (x + y) + \cos (x - y)}}
RHS=tan⁡y = \tan y
Now, we will try to solve the LHS part first, and we will get:
LHS=sin⁡(x+y)−sin⁡(x−y)cos⁡(x+y)+cos⁡(x−y) = \dfrac{{\sin (x + y) - \sin (x - y)}}{{\cos (x + y) + \cos (x - y)}}
Here, we will try to solve the numerator and denominator parts separately.
Here, the numerator is sin⁡(x+y)−sin⁡(x−y)\sin (x + y) - \sin (x - y). We will try to solve the numerator.
We will try to use the basic trigonometric formulas which are:
sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin (A + B) = \sin A\cos B + \cos A\sin B
sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin (A - B) = \sin A\cos B - \cos A\sin B
We will try to put these formulas one by one in our above numerator, and we will get:
sin⁡(x+y)−sin⁡(x−y)=(sin⁡xcos⁡y+cos⁡xsin⁡y)−(sin⁡xcos⁡y−cos⁡xsin⁡y)\sin (x + y) - \sin (x - y) = (\sin x\cos y + \cos x\sin y) - (\sin x\cos y - \cos x\sin y)
Now, we will open the brackets, and we get:
=sin⁡xcos⁡y+cos⁡xsin⁡y−sin⁡xcos⁡y+cos⁡xsin⁡y= \sin x\cos y + \cos x\sin y - \sin x\cos y + \cos x\sin y
Now, we will try to simplify the terms. Some terms get cancel out, and we get:
=cos⁡xsin⁡y+cos⁡xsin⁡y= \cos x\sin y + \cos x\sin y
=2cos⁡xsin⁡y= 2\cos x\sin y
Now, similarly we will try to solve the denominator as well. Here, the denominator is cos⁡(x+y)+cos⁡(x−y)\cos (x + y) + \cos (x - y).
We will try to use the basic trigonometric formulas which are:
cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos (A + B) = \cos A\cos B - \sin A\sin B
cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos (A - B) = \cos A\cos B + \sin A\sin B
We will try to put these formulas one by one in our above denominator, and we will get:
cos⁡(x+y)+cos⁡(x−y)=(cos⁡xcos⁡y−sin⁡xsin⁡y)+(cos⁡xcos⁡y+sin⁡xsin⁡y)\cos (x + y) + \cos (x - y) = (\cos x\cos y - \sin x\sin y) + (\cos x\cos y + \sin x\sin y)
Now, we will open the brackets, and we get:
cos⁡(x+y)+cos⁡(x−y)=cos⁡xcos⁡y−sin⁡xsin⁡y+cos⁡xcos⁡y+sin⁡xsin⁡y\cos (x + y) + \cos (x - y) = \cos x\cos y - \sin x\sin y + \cos x\cos y + \sin x\sin y
Now, we will try to simplify the terms. Some terms get cancel out, and we get:
=cos⁡xcos⁡y+cos⁡xcos⁡y= \cos x\cos y + \cos x\cos y
=2cos⁡xcos⁡y= 2\cos x\cos y
Now, we will put these answers in our LHS part.
LHS=sin⁡(x+y)−sin⁡(x−y)cos⁡(x+y)+cos⁡(x−y) = \dfrac{{\sin (x + y) - \sin (x - y)}}{{\cos (x + y) + \cos (x - y)}}
LHS=2cos⁡xsin⁡y2cos⁡xcos⁡y = \dfrac{{2\cos x\sin y}}{{2\cos x\cos y}}
Here, we can cancel out the like terms, and we get:
LHS=sin⁡ycos⁡y = \dfrac{{\sin y}}{{\cos y}}
We know the basic trigonometric formula tan⁡A=sin⁡Acos⁡A\tan A = \dfrac{{\sin A}}{{\cos A}}. We will put this formula in the LHS part, and we get:
LHS=tan⁡y = \tan y
Here, we can see that LHS=RHS (hence proved). Therefore, our question is solved and proved.

Note: 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 LHS part, we can start solving the RHS part as well. But here we should prefer solving the LHS part only, because it is easy to solve the LHS part.