Prove that: (sin x - sin y)/(cos x + cos y) = tan (x - y)/2 — Mathematics

Question 34 of 50intermediate🔧 ApplyLong Answer4 marks

Prove that:

\frac{\sin x - \sin y}{\cos x + \cos y} = \tan \frac{x - y}{2}

Correct Answer

L.H.S. $= \dfrac{\sin x - \sin y}{\cos x + \cos y}$

Using the sum and difference formulas from the context:

Numerator:

\sin x - \sin y = 2\cos\frac{x+y}{2}\sin\frac{x-y}{2} \quad \text{...[Formula (iv)]}

Denominator:

\cos x + \cos y = 2\cos\frac{x+y}{2}\cos\frac{x-y}{2} \quad \text{...[Formula (i)]}

Therefore,

\text{L.H.S.} = \frac{2\cos\frac{x+y}{2}\sin\frac{x-y}{2}}{2\cos\frac{x+y}{2}\cos\frac{x-y}{2}}

The common factor $2\cos\frac{x+y}{2}$ cancels out from numerator and denominator.

\text{L.H.S.} = \frac{\sin\frac{x-y}{2}}{\cos\frac{x-y}{2}}

\text{L.H.S.} = \tan\frac{x-y}{2}

\text{L.H.S.} = \text{R.H.S.}

∴ Hence Proved.

Exercise: EXERCISE 3.3 | Q: 17 | (Page 25)

For More Understanding

Explanation

This question tests the application of sum-to-product formulas. The key formulas used are:

\sin x - \sin y = 2\cos\frac{x+y}{2}\sin\frac{x-y}{2}

and

\cos x + \cos y = 2\cos\frac{x+y}{2}\cos\frac{x-y}{2}

When we divide these two expressions, the common factor $2\cos\frac{x+y}{2}$ cancels out, leaving $\dfrac{\sin\frac{x-y}{2}}{\cos\frac{x-y}{2}}$ which equals $\tan\frac{x-y}{2}$ .

Solution Steps

Step 1: Write down the L.H.S. expression $\dfrac{\sin x - \sin y}{\cos x + \cos y}$
Step 2: Apply formula (iv): $\sin x - \sin y = 2\cos\dfrac{x+y}{2}\sin\dfrac{x-y}{2}$
Step 3: Apply formula (i): $\cos x + \cos y = 2\cos\dfrac{x+y}{2}\cos\dfrac{x-y}{2}$
Step 4: Divide numerator by denominator and cancel the common factor $2\cos\dfrac{x+y}{2}$
Step 5: Simplify to get $\tan\dfrac{x-y}{2}$ , which equals R.H.S.

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