6.2 Simplify the following expression to ONE trigonometric term: \( \frac{\sin x}{\cos x \cdot \tan x}+\sin \left(180^{\circ}+x\right) \cos \left(90^{\circ}-x\right) \) Consider the identity: \( \frac{\cos 2 \beta}{(\cos \beta+\sin \beta)^{3}}=\frac{\cos \beta-\sin \beta}{1+\sin 2 \beta} \) 6.3.1 Prove the identity: \( \frac{\cos 2 \beta}{(\cos \beta+\sin \beta)^{3}}=\frac{\cos \beta-\sin \beta}{1+\sin 2 \beta} \)

\[ \textbf{6.2 Simplification} \] We start with the expression \[ \frac{\sin x}{\cos x \cdot \tan x}+\sin (180^\circ+x)\cos (90^\circ-x). \] 1. Recognize that \[ \tan x=\frac{\sin x}{\cos x}. \] Therefore, the first term becomes \[ \frac{\sin x}{\cos x\cdot \frac{\sin x}{\cos x}}=\frac{\sin x}{\sin x}=1. \] 2. For the second term, recall the angle transformation formulas: \[ \sin (180^\circ+x)=-\sin x \quad \text{and} \quad \cos (90^\circ-x)=\sin x. \] Thus, \[ \sin(180^\circ+x)\cos(90^\circ-x)=-\sin x\cdot \sin x=-\sin^2 x. \] 3. Combine the results: \[ 1-\sin^2 x. \] Using the Pythagorean identity, \[ 1-\sin^2 x=\cos^2 x. \] The simplified expression is \[ \cos^2 x. \] \[ \textbf{6.3.1 Proof of the Identity} \] We need to prove that \[ \frac{\cos 2\beta}{(\cos \beta+\sin \beta)^{3}}=\frac{\cos \beta-\sin \beta}{1+\sin 2\beta}. \] 1. Begin with the numerator on the left-hand side (LHS). Recall the double-angle formula for cosine: \[ \cos 2\beta=\cos^2\beta-\sin^2\beta=(\cos \beta-\sin \beta)(\cos \beta+\sin \beta). \] Therefore, we can write the LHS as: \[ \frac{(\cos \beta-\sin \beta)(\cos \beta+\sin \beta)}{(\cos \beta+\sin \beta)^3}. \] 2. Simplify the expression by canceling one common factor, \((\cos \beta+\sin \beta)\), in the numerator and denominator: \[ \frac{\cos \beta-\sin \beta}{(\cos \beta+\sin \beta)^2}. \] 3. Next, note that \[ (\cos \beta+\sin \beta)^2=\cos^2\beta+2\sin\beta\cos\beta+\sin^2\beta=1+2\sin\beta\cos\beta. \] Since \[ \sin 2\beta=2\sin\beta\cos\beta, \] we conclude that \[ (\cos \beta+\sin \beta)^2=1+\sin 2\beta. \] 4. Substitute this result into the simplified expression: \[ \frac{\cos \beta-\sin \beta}{1+\sin 2\beta}. \] Thus, we have shown that \[ \frac{\cos 2\beta}{(\cos \beta+\sin \beta)^3}=\frac{\cos \beta-\sin \beta}{1+\sin 2\beta}, \] which completes the proof.