$cos~2x=$

Q: $cos~2x=$

The given question asks for the expression of $ \cos 2x $ in terms of $ \tan x $. The correct answer is provided as Option 1: $ \frac{1-\tan^2 x}{1+\tan^2 x} $. ### Explanation for Option 1: The double angle identity for cosine is given by: \[ \cos 2x = \frac{1 - \tan^2 x}{1 + \tan^2 x} \] To derive this, we start with the basic trigonometric identities: 1. $\tan x = \frac{\sin x}{\cos x}$ 2. $\sin^2 x + \cos^2 x = 1$ Using the double angle identities for sine and cosine: - $\sin 2x = 2 \sin x \cos x$ - $\cos 2x = \cos^2 x - \sin^2 x$ We can express $\cos 2x$ in terms of $\tan x$: \[ \cos 2x = \cos^2 x - \sin^2 x = \left(\frac{\cos^2 x - \sin^2 x}{\cos^2 x}\right) = \frac{1 - \tan^2 x}{1 + \tan^2 x} \] This confirms that Option 1 is indeed the correct expression for $\cos 2x$ in terms of $\tan x$. ### Explanation for Other Options: - **Option 2: $\frac{1+\tan^2 x}{1-\tan^2 x}$** This expression does not correspond to any standard trigonometric identity for $\cos 2x$. In fact, it would lead to undefined values when $\tan x = 1$, which is not consistent with the behavior of $\cos 2x$. - **Option 3: $\frac{2 \tan x}{1+\tan^2 x}$** This expression is actually the identity for $\sin 2x$, not $\cos 2x$. Therefore, it is incorrect for representing $\cos 2x$. - **Option 4: $\frac{1-\tan^2 x}{2 \tan x}$** This expression does not match any known trigonometric identity for $\cos 2x$ and would not simplify to a valid form for $\cos 2x$. In conclusion, Option 1 is the correct expression for $\cos 2x$ in terms of $\tan x$, and the other options do not accurately represent this trigonometric identity.

Question

$cos~2x=$

Accepted Answer

The given question asks for the expression of $ \cos 2x $ in terms of $ \tan x $. The correct answer is provided as Option 1: $ \frac{1-\tan^2 x}{1+\tan^2 x} $.

### Explanation for Option 1:

The double angle identity for cosine is given by:

$$
\cos 2x = \frac{1 - \tan^2 x}{1 + \tan^2 x}
$$

To derive this, we start with the basic trigonometric identities:

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

Using the double angle identities for sine and cosine:

- $\sin 2x = 2 \sin x \cos x$
- $\cos 2x = \cos^2 x - \sin^2 x$

We can express $\cos 2x$ in terms of $\tan x$:

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

This confirms that Option 1 is indeed the correct expression for $\cos 2x$ in terms of $\tan x$.

### Explanation for Other Options:

- **Option 2: $\frac{1+\tan^2 x}{1-\tan^2 x}$**

This expression does not correspond to any standard trigonometric identity for $\cos 2x$. In fact, it would lead to undefined values when $\tan x = 1$, which is not consistent with the behavior of $\cos 2x$.

- **Option 3: $\frac{2 \tan x}{1+\tan^2 x}$**

This expression is actually the identity for $\sin 2x$, not $\cos 2x$. Therefore, it is incorrect for representing $\cos 2x$.

- **Option 4: $\frac{1-\tan^2 x}{2 \tan x}$**

This expression does not match any known trigonometric identity for $\cos 2x$ and would not simplify to a valid form for $\cos 2x$.

In conclusion, Option 1 is the correct expression for $\cos 2x$ in terms of $\tan x$, and the other options do not accurately represent this trigonometric identity.