Prove that: x 2x - 2x 3x - 3x x = 1

[object Object] — The problem requires proving a trigonometric identity involving cotangent functions of multiples of x. The key strategy is to recognize that the term 3x can be expanded using the compound angle formula (A+B). The provided context explicitly lists the identity: (x + y) = x y - 1 y + x By applying this identity to (2x+x), we derive a relationship between 3x, 2x, and x. Substituting this relationship back into the original expression simplifies the terms to yield the result 1 .

Prove that: $$\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = 1$$

Question 21 of 50advanced🔧 ApplyLong Answer4 marks

Prove that:

$\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = 1$

Correct Answer

We need to prove that $\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = 1$ .

Let us consider the L.H.S.:

$\text{L.H.S.} = \cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x$

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

$= \cot x \cot 2x - \cot 3x (\cot 2x + \cot x)$

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

$\cot (A + B) = \frac{\cot A \cot B - 1}{\cot B + \cot A}$

Substituting $A = 2x$ and $B = x$ , we get:

$\cot (2x + x) = \frac{\cot 2x \cot x - 1}{\cot x + \cot 2x}$

This can be rewritten as:

$\cot 3x = \frac{\cot 2x \cot x - 1}{\cot x + \cot 2x}$

Cross-multiplying gives:

$\cot 3x (\cot x + \cot 2x) = \cot 2x \cot x - 1$

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

$\text{L.H.S.} = \cot x \cot 2x - (\cot 2x \cot x - 1)$

$= \cot x \cot 2x - \cot x \cot 2x + 1$

$= 1$

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

Exercise: EXERCISE 3.3 | Q: 22 | (Chapter: 26)

For More Understanding

Explanation

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

The provided context explicitly lists the identity:

$\cot (x + y) = \frac{\cot x \cot y - 1}{\cot y + \cot x}$

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

Solution Steps

Step 1: Rearrange the L.H.S. expression $\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x$ as $\cot x \cot 2x - \cot 3x(\cot 2x + \cot x)$ .
Step 2: Use the identity $\cot(A+B) = \frac{\cot A \cot B - 1}{\cot B + \cot A}$ from the context on the term $\cot 3x$ (where $3x = 2x + x$ ).
Step 3: Apply the identity to get $\cot 3x = \frac{\cot 2x \cot x - 1}{\cot x + \cot 2x}$ , which implies $\cot 3x(\cot x + \cot 2x) = \cot 2x \cot x - 1$ .
Step 4: Substitute $(\cot 2x \cot x - 1)$ for $\cot 3x(\cot 2x + \cot x)$ in the rearranged L.H.S. expression.
Step 5: Simplify the expression $\cot x \cot 2x - (\cot x \cot 2x - 1)$ to get 1, thus proving L.H.S. = R.H.S.

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