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:
We must expand the square on the left-hand side and simplify it using trigonometric identities to match the format on the right-hand side.
Step 2: Key Formula or Approach:
Use \((a - b)^2 = a^2 - 2ab + b^2\).
Use \(\cos x \cdot \sec x = 1\).
Use the identity \(\sec^2 x = 1 + \tan^2 x\).
Step 3: Detailed Explanation:
Expand the left-hand side (LHS):
\[ (3\cos x - 2\sec x)^2 = (3\cos x)^2 - 2(3\cos x)(2\sec x) + (2\sec x)^2 \] \[ = 9\cos^2 x - 12(\cos x \cdot \sec x) + 4\sec^2 x \] Since \(\cos x \cdot \sec x = 1\):
\[ = 9\cos^2 x - 12 + 4\sec^2 x \] We need to introduce \(\tan^2 x\) to match the right-hand side (RHS). Substitute \(\sec^2 x = 1 + \tan^2 x\):
\[ = 9\cos^2 x - 12 + 4(1 + \tan^2 x) \] \[ = 9\cos^2 x - 12 + 4 + 4\tan^2 x \] \[ = 9\cos^2 x + 4\tan^2 x - 8 \] Equate this to the RHS:
\[ 9\cos^2 x + 4\tan^2 x - 8 = 9\cos^2 x + 4\tan^2 x + k \] By comparison, \(k = -8\).
Step 4: Final Answer:
The value of \(k\) is -8.