d. $$ \frac{1-\cos 2 heta}{1+\cos 2 heta}= an ^{2} heta $$

Question

d. $$ \frac{1-\cos 2 	heta}{1+\cos 2 	heta}=	an ^{2} 	heta $$

UpStudy AI · Accepted Answer

To prove the identity

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

we can start by using the double angle identity for cosine:

$$
\cos 2\theta = 1 - 2\sin^2 \theta.
$$

Substituting this into the left-hand side, we have:

$$
1 - \cos 2\theta = 1 - (1 - 2\sin^2 \theta) = 2\sin^2 \theta,
$$

and

$$
1 + \cos 2\theta = 1 + (1 - 2\sin^2 \theta) = 2 - 2\sin^2 \theta = 2(1 - \sin^2 \theta) = 2\cos^2 \theta.
$$

Now substituting these results back into the left-hand side of the original equation gives:

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

Thus, we have shown that:

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

Therefore, the identity is proven to be true.
The identity $$ \frac{1 - \cos 2\theta}{1 + \cos 2\theta} = \tan^2 \theta $$ is true.

d. \( \frac{1-\cos 2 \theta}{1+\cos 2 \theta}=\tan ^{2} \theta \)