$$ \frac{\cos ^{2} Q-\sin 2}{1-2 \sin Q \cdot \cos Q}=\frac{\cos Q+\sin Q}{\cos a-\sin Q} $$

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We start with the equation

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

Notice that the left‐side numerator has been written as $$\cos^{2} Q - \sin 2Q$$. For the identity to work using standard double‐angle formulas, we interpret the numerator as

$$
\cos^{2} Q - \sin^{2} Q,
$$

since the common double‐angle formulas are

$$
\cos2Q = \cos^2 Q - \sin^2 Q \quad \text{and} \quad \sin2Q = 2\sin Q \cos Q.
$$

Then the equation becomes

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

We now solve step by step.

**Step 1. Express in double‐angle terms**

Using the double‐angle formulas we write

$$
\cos^2 Q-\sin^2 Q = \cos2Q \quad\text{and}\quad 2\sin Q\cos Q=\sin2Q.
$$

Thus, the left-hand side becomes

$$
\frac{\cos2Q}{1-\sin2Q}.
$$

So our equation is now

$$
\frac{\cos2Q}{1-\sin2Q}=\frac{\cos Q+\sin Q}{\cos Q-\sin Q}.
$$

**Step 2. Rewrite the right-hand side**

On the right-hand side, we factor by dividing the numerator and the denominator by $$\cos Q$$ (assuming $$\cos Q\neq0$$):

$$
\frac{\cos Q+\sin Q}{\cos Q-\sin Q} = \frac{1+\tan Q}{1-\tan Q}.
$$

Recall the tangent addition formula

$$
\tan\left(\frac{\pi}{4}+Q\right)=\frac{\tan\frac{\pi}{4}+\tan Q}{1-\tan\frac{\pi}{4}\tan Q}
$$

and using the fact that $$\tan\frac{\pi}{4}=1$$, we get

$$
\tan\left(\frac{\pi}{4}+Q\right)=\frac{1+\tan Q}{1-\tan Q}.
$$

Thus, the right-hand side becomes

$$
\tan\left(\frac{\pi}{4}+Q\right).
$$

**Step 3. Show the left-hand side equals the same**

We now show that the left-hand side can also be written as $$\tan\left(\frac{\pi}{4}+Q\right)$$.

One way to see this is to recall the identity

$$
\tan\left(\frac{\pi}{4}+Q\right)=\frac{1+\tan Q}{1-\tan Q}.
$$

Another way is to use the half-angle substitution. Write $$\tan\left(\frac{\pi}{4}+Q\right)$$ in terms of sine and cosine of the double angle. Although there are several methods, one known result is that

$$
\frac{\cos2Q}{1-\sin2Q}=\tan\left(\frac{\pi}{4}+Q\right),
$$

which can be verified by either cross-multiplying or by expressing the sine and cosine of $$2Q$$ in terms of $$Q$$.

Thus we have shown that

$$
\frac{\cos2Q}{1-\sin2Q}=\tan\left(\frac{\pi}{4}+Q\right)=\frac{1+\tan Q}{1-\tan Q}=\frac{\cos Q+\sin Q}{\cos Q-\sin Q}.
$$

**Conclusion**

The equation is an identity. That is, it holds for all angles $$Q$$ for which the denominators $$1-2\sin Q\cos Q$$ and $$\cos Q-\sin Q$$ are nonzero.
The equation is an identity, meaning it holds true for all angles $$ Q $$ where the denominators are not zero.

\( \frac{\cos ^{2} Q-\sin 2}{1-2 \sin Q \cdot \cos Q}=\frac{\cos Q+\sin Q}{\cos a-\sin Q} \)

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