Question

\(\frac{1-\cos 2x}{1+\cos 2x}-\sec^2 x=\)

• A. \(1\)
• B. \(\tan 2x\)
• C. \(\sec 2x\)
• D. \(0\)
• E. \(-1\)

Hint:

Memorize identities involving double angles and \(\sec^2 x=1+\tan^2 x\). They help simplify expressions quickly.

Answer 1

Answer: (E)

Step 1: Start with the given expression.
\[ \frac{1-\cos 2x}{1+\cos 2x}-\sec^2 x \]

Step 2: Use trigonometric identities.
Recall: \[ 1-\cos 2x=2\sin^2 x \] \[ 1+\cos 2x=2\cos^2 x \]

Step 3: Substitute into the expression.
\[ \frac{2\sin^2 x}{2\cos^2 x}-\sec^2 x \]

Step 4: Simplify the fraction.
\[ \frac{2\sin^2 x}{2\cos^2 x}=\frac{\sin^2 x}{\cos^2 x}=\tan^2 x \] So the expression becomes: \[ \tan^2 x-\sec^2 x \]

Step 5: Use identity relating \(\sec^2 x\) and \(\tan^2 x\).
We know: \[ \sec^2 x=1+\tan^2 x \]

Step 6: Substitute and simplify.
\[ \tan^2 x-(1+\tan^2 x)=-1 \]

Step 7: Final answer.
Thus, the value is: \[ \boxed{-1} \] which matches option \((5)\).