Q.Prove that the determinant $\begin{vmatrix} x & \sin\theta & \cos\theta \\ -\sin\theta & -x & 1 \\ \cos\theta & 1 & x \end{vmatrix}$ is independent of $\theta$.

Concept: Determinant Evaluation Using Identities — we expand and simplify using $\sin^2\theta + \cos^2\theta = 1$ to show the $\theta$ terms cancel.

Step 1: Expand the determinant along the first row:

Step 2: Compute each $2\times2$ determinant:

Step 3: Substitute back:

Step 4: The terms $\sin\theta\cos\theta$ cancel. Using $\sin^2\theta + \cos^2\theta = 1$:

The key idea is to treat the determinant as an expression in $\theta$ and see if it actually depends on $\theta$ at all. Often, determinants with trigonometric entries simplify using identities like $\sin^2\theta + \cos^2\theta = 1$, or by expanding and grouping terms. Here, a direct expansion will work cleanly — no row operations needed.

Let’s go step by step.

1. Write the determinant We have

2. Expand along the first row (or any row — first row is fine because it has $x$, $\sin\theta$, $\cos\theta$). Using the standard formula for a $3\times3$ determinant:

3. Compute each $2\times2$ determinant
   • First minor:

• Second minor:

• Third minor:

4. Substitute back into the expansion

Simplify term by term:

• First term: $x(-x^2 - 1) = -x^3 - x$.
• Second term: $-\sin\theta(-x\sin\theta - \cos\theta) = \sin\theta \cdot (x\sin\theta + \cos\theta) = x\sin^2\theta + \sin\theta\cos\theta$.
• Third term: $\cos\theta(-\sin\theta + x\cos\theta) = -\sin\theta\cos\theta + x\cos^2\theta$.

5. Combine everything

Notice $\sin\theta\cos\theta$ and $-\sin\theta\cos\theta$ cancel each other exactly.

So we are left with:

6. Use the Pythagorean identity

Hence,

Therefore,

The final expression contains no $\theta$ at all — it is simply $-x^3$, a function of $x$ alone. So the determinant is independent of $\theta$.

Method: Proving a Determinant Is Independent of a Parameter (e.g. $\theta$)

When a question asks you to show a determinant does NOT depend on some angle or variable, the strategy is to expand it fully and show every occurrence of that variable cancels out algebraically.

Steps

Step 1: Expand the determinant along the row or column that looks simplest

Choose the row/column with the fewest or simplest trigonometric entries to minimise the number of terms you carry forward.

Step 2: Compute every $2\times2$ minor carefully, keeping the trig terms unexpanded

Write out each minor as a product/difference of sines and cosines without simplifying yet — premature simplification is where sign errors creep in.

Step 3: Substitute the minors back and collect like terms

Group terms that are pure functions of the "other" variable (here $x$) separately from terms that still carry the parameter (here $\theta$).

Step 4: Use a trigonometric identity to eliminate the parameter

Look for a combination like $\sin^2\theta + \cos^2\theta$ hiding in the collected terms — replacing it with $1$ is usually what makes the parameter vanish and confirms independence. If a $\sin\theta\cos\theta$ term appears twice with opposite signs, note that it cancels directly without needing any identity.

Common Mistakes

Mistake 1: Expanding along a row that leaves the messiest arithmetic

Why it's wrong: some rows/columns lead to far more terms to track than others; picking a "hard" row makes it much easier to drop a sign or a term. Correct approach: scan all three rows/columns first and expand along the one with the fewest distinct trig products.

Mistake 2: Missing the cancelling $\sin\theta\cos\theta$ terms

Why it's wrong: in problems like this, two cross-terms with opposite signs cancel exactly — if you simplify too aggressively or too early, it's easy to lose track of one of them and end up with a leftover $\theta$-term that shouldn't be there. Correct approach: keep all terms explicit until the very end, then cancel matching pairs deliberately, one at a time.

Mistake 3: Forgetting to apply $\sin^2\theta+\cos^2\theta=1$ to finish the proof

Why it's wrong: stopping right after collecting terms like $x\sin^2\theta + x\cos^2\theta$ without simplifying them to $x$ leaves the expression looking like it still depends on $\theta$, even though it doesn't. Correct approach: always scan the final expression for a $\sin^2+\cos^2$ pattern and apply the identity before declaring the proof complete.