Question 37 of 50advanced🔧 ApplyLong Answer4 marks

Prove that:

cot⁡xcot⁡2x−cot⁡2xcot⁡3x−cot⁡3xcot⁡x=1\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = 1
Correct Answer

We need to prove that cot⁡xcot⁡2x−cot⁡2xcot⁡3x−cot⁡3xcot⁡x=1\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = 1.

Let us consider the L.H.S.:

L.H.S.=cot⁡xcot⁡2x−cot⁡2xcot⁡3x−cot⁡3xcot⁡x\text{L.H.S.} = \cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x

We can rearrange the terms by taking −cot⁡3x-\cot 3x common from the last two terms:

=cot⁡xcot⁡2x−cot⁡3x(cot⁡2x+cot⁡x)= \cot x \cot 2x - \cot 3x (\cot 2x + \cot x)

We know that 3x=2x+x3x = 2x + x. Using the identity for the cotangent of a sum provided in the text:

cot⁡(A+B)=cot⁡Acot⁡B−1cot⁡B+cot⁡A\cot (A + B) = \frac{\cot A \cot B - 1}{\cot B + \cot A}

Substituting A=2xA = 2x and B=xB = x, we get:

cot⁡(2x+x)=cot⁡2xcot⁡x−1cot⁡x+cot⁡2x\cot (2x + x) = \frac{\cot 2x \cot x - 1}{\cot x + \cot 2x}

This can be rewritten as:

cot⁡3x=cot⁡2xcot⁡x−1cot⁡x+cot⁡2x\cot 3x = \frac{\cot 2x \cot x - 1}{\cot x + \cot 2x}

Cross-multiplying gives:

cot⁡3x(cot⁡x+cot⁡2x)=cot⁡2xcot⁡x−1\cot 3x (\cot x + \cot 2x) = \cot 2x \cot x - 1

Now, we substitute this value back into the expression for L.H.S.:

L.H.S.=cot⁡xcot⁡2x−(cot⁡2xcot⁡x−1)\text{L.H.S.} = \cot x \cot 2x - (\cot 2x \cot x - 1) =cot⁡xcot⁡2x−cot⁡xcot⁡2x+1= \cot x \cot 2x - \cot x \cot 2x + 1 =1= 1

Thus, L.H.S. = R.H.S. Hence proved.

Exercise: EXERCISE 3.3 | Q: 22 | (Page 26)
For More Understanding

Explanation

The problem requires proving a trigonometric identity involving cotangent functions of multiples of xx. The key strategy is to recognize that the term cot⁡3x\cot 3x can be expanded using the compound angle formula cot⁡(A+B)\cot(A+B).

The provided context explicitly lists the identity:

cot⁡(x+y)=cot⁡xcot⁡y−1cot⁡y+cot⁡x\cot (x + y) = \frac{\cot x \cot y - 1}{\cot y + \cot x}

By applying this identity to cot⁡(2x+x)\cot(2x+x), we derive a relationship between cot⁡3x\cot 3x, cot⁡2x\cot 2x, and cot⁡x\cot x. Substituting this relationship back into the original expression simplifies the terms to yield the result 1.

Solution Steps

  1. Step 1: Rearrange the L.H.S. expression cot⁡xcot⁡2x−cot⁡2xcot⁡3x−cot⁡3xcot⁡x\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x as cot⁡xcot⁡2x−cot⁡3x(cot⁡2x+cot⁡x)\cot x \cot 2x - \cot 3x(\cot 2x + \cot x).

  2. Step 2: Use the identity cot⁡(A+B)=cot⁡Acot⁡B−1cot⁡B+cot⁡A\cot(A+B) = \frac{\cot A \cot B - 1}{\cot B + \cot A} from the context on the term cot⁡3x\cot 3x (where 3x=2x+x3x = 2x + x).

  3. Step 3: Apply the identity to get cot⁡3x=cot⁡2xcot⁡x−1cot⁡x+cot⁡2x\cot 3x = \frac{\cot 2x \cot x - 1}{\cot x + \cot 2x}, which implies cot⁡3x(cot⁡x+cot⁡2x)=cot⁡2xcot⁡x−1\cot 3x(\cot x + \cot 2x) = \cot 2x \cot x - 1.

  4. Step 4: Substitute (cot⁡2xcot⁡x−1)(\cot 2x \cot x - 1) for cot⁡3x(cot⁡2x+cot⁡x)\cot 3x(\cot 2x + \cot x) in the rearranged L.H.S. expression.

  5. Step 5: Simplify the expression cot⁡xcot⁡2x−(cot⁡xcot⁡2x−1)\cot x \cot 2x - (\cot x \cot 2x - 1) to get 1, thus proving L.H.S. = R.H.S.

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