Question

If $(3 \cos x - 2 \sec x)^2 = 9 \cos^2 x + 4 \tan^2 x + k$, where $k$ is a constant, then the value of $k$ is equal to

• A. 12
• B. -12
• C. -4
• D. 8
• E. -8

Hint:

For identities where $k$ is a constant, you can pick a specific value for $x$ (like $x=0$) to solve quickly.
If $x=0$, $(3(1) - 2(1))^2 = 9(1)^2 + 4(0)^2 + k \implies 1^2 = 9 + k \implies k = -8$.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
Expand the left-hand side of the identity and use basic trigonometric identities like $\sec x \cdot \cos x = 1$ and $\sec^2 x = 1 + \tan^2 x$ to match the form on the right-hand side.

Step 2: Detailed Explanation:
1. Expand the left-hand side $(a - b)^2 = a^2 + b^2 - 2ab$:
\[ (3\cos x - 2\sec x)^2 = (3\cos x)^2 + (2\sec x)^2 - 2(3\cos x)(2\sec x) \]
\[ = 9\cos^2 x + 4\sec^2 x - 12(\cos x \sec x) \]
2. Use the identity $\cos x \sec x = 1$:
\[ = 9\cos^2 x + 4\sec^2 x - 12 \]
3. Substitute $\sec^2 x = 1 + \tan^2 x$:
\[ = 9\cos^2 x + 4(1 + \tan^2 x) - 12 \]
\[ = 9\cos^2 x + 4 + 4\tan^2 x - 12 \]
4. Combine constant terms:
\[ = 9\cos^2 x + 4\tan^2 x - 8 \]
5. Compare this with the given RHS: $9 \cos^2 x + 4 \tan^2 x + k$.
Clearly, $k = -8$.

Step 3: Final Answer:
The value of $k$ is -8.