Question

The value of

$F(x) = 6\cos x\sqrt{1+\tan^2x} + 2\sin x\sqrt{1+\cot^2x}$,

Where $x \in (0, 2\pi) - \{\pi, \frac{\pi}{2}, \frac{3\pi}{2}\}$ may be

• A. 4
• B. -4
• C. 8
• D. -8

Answer 1

Answer: (A), (B), (C), (D)

All options are correct since F(x) can be 4, -4, 8, or -8 depending on the quadrant of x.

Answer 2

The given function is $F(x) = 6\cos x\sqrt{1+\tan^2x} + 2\sin x\sqrt{1+\cot^2x}$.

We know the trigonometric identities: $1+\tan^2x = \sec^2x$ $1+\cot^2x = \csc^2x$

Substituting these into the expression for $F(x)$: $F(x) = 6\cos x\sqrt{\sec^2x} + 2\sin x\sqrt{\csc^2x}$

Using the property $\sqrt{A^2} = |A|$, we get: $F(x) = 6\cos x|\sec x| + 2\sin x|\csc x|$

We know that $\sec x = \frac{1}{\cos x}$ and $\csc x = \frac{1}{\sin x}$. The domain given is $x \in (0, 2\pi) - \{\pi, \frac{\pi}{2}, \frac{3\pi}{2}\}$. This means $\cos x \neq 0$ and $\sin x \neq 0$.

We need to consider the signs of $\cos x$ and $\sin x$ in different quadrants to evaluate the absolute values.

Term 1: $6\cos x|\sec x|$

• If $\cos x > 0$ (i.e., $x$ is in Quadrant I or Quadrant IV), then $\sec x > 0$, so $|\sec x| = \sec x$. In this case, $6\cos x|\sec x| = 6\cos x\left(\frac{1}{\cos x}\right) = 6$.
• If $\cos x < 0$ (i.e., $x$ is in Quadrant II or Quadrant III), then $\sec x < 0$, so $|\sec x| = -\sec x$. In this case, $6\cos x|\sec x| = 6\cos x\left(-\frac{1}{\cos x}\right) = -6$.

Term 2: $2\sin x|\csc x|$

• If $\sin x > 0$ (i.e., $x$ is in Quadrant I or Quadrant II), then $\csc x > 0$, so $|\csc x| = \csc x$. In this case, $2\sin x|\csc x| = 2\sin x\left(\frac{1}{\sin x}\right) = 2$.
• If $\sin x < 0$ (i.e., $x$ is in Quadrant III or Quadrant IV), then $\csc x < 0$, so $|\csc x| = -\csc x$. In this case, $2\sin x|\csc x| = 2\sin x\left(-\frac{1}{\sin x}\right) = -2$.

Now, let's combine these possibilities based on the quadrant $x$ lies in:

1. If $x \in (0, \frac{\pi}{2})$ (Quadrant I): $\cos x > 0$ and $\sin x > 0$. $F(x) = 6 + 2 = 8$.

2. If $x \in (\frac{\pi}{2}, \pi)$ (Quadrant II): $\cos x < 0$ and $\sin x > 0$. $F(x) = -6 + 2 = -4$.

3. If $x \in (\pi, \frac{3\pi}{2})$ (Quadrant III): $\cos x < 0$ and $\sin x < 0$. $F(x) = -6 - 2 = -8$.

4. If $x \in (\frac{3\pi}{2}, 2\pi)$ (Quadrant IV): $\cos x > 0$ and $\sin x < 0$. $F(x) = 6 - 2 = 4$.

Thus, the possible values of $F(x)$ are $8, -4, -8, 4$. All the given options (A) 4, (B) -4, (C) 8, and (D) -8 are possible values for $F(x)$.