Answer
The equation is an identity, meaning it holds true for all angles \( Q \) where the denominators are not zero.
Solution
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.
Answered by UpStudy AI and reviewed by a Professional Tutor

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