If ` x in ( - pi/2, pi/2)`, then the value of `tan^(-1) ((tan x)/4) + tan^(-1) ((3 sin 2x)/(5 + 3 cos 2x)) ` is

Options

A. `x//2`
B. `2x`
C. `3x`
D. `x`

Answer

To solve the problem, we need to evaluate the expression:

\[ \tan^{-1}\left(\frac{\tan x}{4}\right) + \tan^{-1}\left(\frac{3 \sin 2x}{5 + 3 \cos 2x}\right) \]

**Step 1: Use the double angle identities for sine and cosine.**

Recall that:
\[ \sin 2x = 2 \sin x \cos x \]
\[ \cos 2x = 2 \cos^2 x - 1 \]

Substituting these into the second term gives us:
\[ \tan^{-1}\left(\frac{3 \cdot 2 \sin x \cos x}{5 + 3(2 \cos^2 x - 1)}\right) = \tan^{-1}\left(\frac{6 \sin x \cos x}{5 + 6 \cos^2 x - 3}\right) = \tan^{-1}\left(\frac{6 \sin x \cos x}{2 + 6 \cos^2 x}\right) \]

**Step 2: Simplify the expression.**

Now we can rewrite the expression as:
\[ \tan^{-1}\left(\frac{\tan x}{4}\right) + \tan^{-1}\left(\frac{6 \sin x \cos x}{2 + 6 \cos^2 x}\right) \]

**Step 3: Rewrite the second term.**

Notice that:
\[ \frac{6 \sin x \cos x}{2 + 6 \cos^2 x} = \frac{6 \tan x \cos^2 x}{2 + 6 \cos^2 x} \]

**Step 4: Use the formula for the sum of arctangents.**

We can use the formula:
\[ \tan^{-1} a + \tan^{-1} b = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \]
where \( a = \frac{\tan x}{4} \) and \( b = \frac{6 \tan x \cos^2 x}{2 + 6 \cos^2 x} \).

**Step 5: Calculate \( a + b \) and \( 1 - ab \).**

First, calculate \( a + b \):
\[ a + b = \frac{\tan x}{4} + \frac{6 \tan x \cos^2 x}{2 + 6 \cos^2 x} \]

Now, calculate \( ab \):
\[ ab = \left(\frac{\tan x}{4}\right) \left(\frac{6 \tan x \cos^2 x}{2 + 6 \cos^2 x}\right) = \frac{6 \tan^2 x \cos^2 x}{4(2 + 6 \cos^2 x)} \]

**Step 6: Substitute into the formula.**

Now we substitute \( a + b \) and \( 1 - ab \) into the formula:
\[ \tan^{-1}\left(\frac{\frac{\tan x}{4} + \frac{6 \tan x \cos^2 x}{2 + 6 \cos^2 x}}{1 - \frac{6 \tan^2 x \cos^2 x}{4(2 + 6 \cos^2 x)}}\right) \]

**Step 7: Simplify the expression.**

After simplification, we find that the expression simplifies to \( x \):
\[ \tan^{-1}(\tan x) = x \quad \text{(since \( x \in (-\frac{\pi}{2}, \frac{\pi}{2}) \))} \]

Thus, the final answer is:
\[ \boxed{x} \]

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