$$ \begin{aligned} ext { Given: } & \frac{2 an \beta}{1+ an ^{2} \beta} \ ext { 5.4.1 } \quad & ext { Prove that: } \ & \frac{2 an \beta}{1+ an ^{2} \beta}=\sin 2 \beta \ ext { 3.4.2 } & ext { Hence, or otherwise determine the maximum value of: } \ & \frac{(1+ an \beta)^{2}}{1+ an ^{2} \beta}\end{aligned} $$

Question

$$ \begin{aligned} 	ext { Given: } & \frac{2 	an \beta}{1+	an ^{2} \beta} \ 	ext { 5.4.1 } \quad & 	ext { Prove that: } \ & \frac{2 	an \beta}{1+	an ^{2} \beta}=\sin 2 \beta \ 	ext { 3.4.2 } & 	ext { Hence, or otherwise determine the maximum value of: } \ & \frac{(1+	an \beta)^{2}}{1+	an ^{2} \beta}\end{aligned} $$

UpStudy AI · Accepted Answer

To solve the problem, we will follow the steps outlined in the prompt.

### Step 1: Prove that $$ \frac{2 \tan \beta}{1+\tan^2 \beta} = \sin 2\beta $$

We start with the left-hand side:

$$
\frac{2 \tan \beta}{1+\tan^2 \beta}
$$

Using the double angle identity for sine, we know that:

$$
\sin 2\beta = \frac{2 \sin \beta \cos \beta}{1}
$$

We can express $$\tan \beta$$ in terms of $$\sin$$ and $$\cos$$:

$$
\tan \beta = \frac{\sin \beta}{\cos \beta}
$$

Substituting this into our expression, we have:

$$
\frac{2 \tan \beta}{1+\tan^2 \beta} = \frac{2 \frac{\sin \beta}{\cos \beta}}{1 + \left(\frac{\sin \beta}{\cos \beta}\right)^2}
$$

Now, simplifying the denominator:

$$
1 + \tan^2 \beta = 1 + \frac{\sin^2 \beta}{\cos^2 \beta} = \frac{\cos^2 \beta + \sin^2 \beta}{\cos^2 \beta} = \frac{1}{\cos^2 \beta}
$$

Thus, we can rewrite our expression:

$$
\frac{2 \frac{\sin \beta}{\cos \beta}}{\frac{1}{\cos^2 \beta}} = 2 \sin \beta \cos \beta
$$

Now, we recognize that:

$$
2 \sin \beta \cos \beta = \sin 2\beta
$$

This proves that:

$$
\frac{2 \tan \beta}{1+\tan^2 \beta} = \sin 2\beta
$$

### Step 2: Determine the maximum value of $$ \frac{(1+\tan \beta)^{2}}{1+\tan^{2} \beta} $$

Let $$ x = \tan \beta $$. We need to find the maximum value of:

$$
f(x) = \frac{(1+x)^2}{1+x^2}
$$

Expanding the numerator:

$$
f(x) = \frac{1 + 2x + x^2}{1 + x^2}
$$

This can be separated into two parts:

$$
f(x) = 1 + \frac{2x}{1+x^2}
$$

Now, we need to find the maximum of $$ \frac{2x}{1+x^2} $$. To do this, we can take the derivative and set it to zero:

Let $$ g(x) = \frac{2x}{1+x^2} $$.

Using the quotient rule:

$$
g'(x) = \frac{(1+x^2)(2) - 2x(2x)}{(1+x^2)^2} = \frac{2 + 2x^2 - 4x^2}{(1+x^2)^2} = \frac{2 - 2x^2}{(1+x^2)^2}
$$

Setting the numerator equal to zero:

$$
2 - 2x^2 = 0 \implies x^2 = 1 \implies x = 1 \text{ (since } x = \tan \beta \text{ must be non-negative)}
$$

Now, we evaluate $$ g(x) $$ at $$ x = 1 $$:

$$
g(1) = \frac{2(1)}{1+1^2} = \frac{2}{2} = 1
$$

Thus, the maximum value of $$ f(x) $$ is:

$$
f(1) = 1 + 1 = 2
$$

### Conclusion

The maximum value of $$ \frac{(1+\tan \beta)^{2}}{1+\tan^{2} \beta} $$ is:

$$
\boxed{2}
$$
The maximum value of $$ \frac{(1+\tan \beta)^{2}}{1+\tan^{2} \beta} $$ is 2.

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