(1) \( 1+\frac{2 \cos x(1+\cos x)}{\sin ^{2} x}=\frac{1+\cos x}{1-\cos x} \) f) \( \frac{1-2 \sin \theta \cos \theta}{\sin ^{2} \theta-\cos ^{2} \theta}=\frac{\sin \theta-\cos \theta}{\sin \theta+\cos \theta} \) g) \( \frac{2 \cos ^{2} \theta+\cos \theta-1}{\sin ^{2} \theta}=\frac{2 \cos \theta-1}{1-\cos \theta} \) h) \( \frac{2 \cos ^{2} \beta+5 \cos \beta-3}{2 \sin \beta \cos \beta-\sin \beta}=\frac{\cos \beta+3}{\sin \beta} \)

(1) We start with \[ 1+\frac{2\cos x\,(1+\cos x)}{\sin^2 x}=\frac{1+\cos x}{1-\cos x}. \] A common method is to eliminate the denominators. Multiply both sides by \(\sin^2 x(1-\cos x)\) (assuming \(\sin x\ne0\) and \(\cos x\ne1\)): \[ \sin^2 x(1-\cos x)+2\cos x\,(1+\cos x)(1-\cos x) =\sin^2 x(1+\cos x). \] Notice that \[ (1+\cos x)(1-\cos x)=1-\cos^2 x=\sin^2 x. \] Thus the left‐hand side becomes \[ \sin^2 x(1-\cos x)+2\cos x\;\sin^2 x=\sin^2 x\,(1-\cos x+2\cos x) =\sin^2 x\,(1+\cos x). \] Since the right‐hand side is also \(\sin^2 x\,(1+\cos x)\), the equality is an identity. (Remember that we must require \(\sin x\ne0\) and \(\cos x\ne1\) so that no denominator is zero.) --- f) We consider \[ \frac{1-2\sin\theta\cos\theta}{\sin^2\theta-\cos^2\theta} =\frac{\sin\theta-\cos\theta}{\sin\theta+\cos\theta}. \] Write the numerator on the left as \[ 1-2\sin\theta\cos\theta=\sin^2\theta+\cos^2\theta-2\sin\theta\cos\theta =(\sin\theta-\cos\theta)^2. \] Also, factor the denominator: \[ \sin^2\theta-\cos^2\theta = (\sin\theta-\cos\theta)(\sin\theta+\cos\theta). \] Thus, for \(\sin\theta-\cos\theta\ne0\) (and also \(\sin\theta+\cos\theta\ne0\)) we have \[ \frac{(\sin\theta-\cos\theta)^2}{(\sin\theta-\cos\theta)(\sin\theta+\cos\theta)} =\frac{\sin\theta-\cos\theta}{\sin\theta+\cos\theta}. \] So the equality holds (with the understanding that when \(\sin\theta-\cos\theta=0\) the original expression must be interpreted as a removable case provided the overall expression is defined). --- g) Start with \[ \frac{2\cos^2\theta+\cos\theta-1}{\sin^2\theta}=\frac{2\cos\theta-1}{1-\cos\theta}. \] Factor the numerator on the left. Notice that \[ (2\cos\theta-1)(\cos\theta+1)=2\cos^2\theta+2\cos\theta-\cos\theta-1 =2\cos^2\theta+\cos\theta-1. \] Then the left‐hand side becomes \[ \frac{(2\cos\theta-1)(\cos\theta+1)}{\sin^2\theta}. \] Assume \(2\cos\theta-1\ne0\) and cancel it (later one should check the possibility \(\cos\theta=\frac12\) separately): \[ \frac{\cos\theta+1}{\sin^2\theta}=\frac{1}{1-\cos\theta}. \] Cross-multiply: \[ (\cos\theta+1)(1-\cos\theta)=\sin^2\theta. \] But \[ (\cos\theta+1)(1-\cos\theta)=1-\cos^2\theta=\sin^2\theta. \] Thus the equality holds. (One must also have \(\sin\theta\ne0\) and \(1-\cos\theta\ne0\) so that no denominator vanishes; when \(2\cos\theta-1=0\) the original expression is interpreted by continuity.) --- h) We have \[ \frac{2\cos^2\beta+5\cos\beta-3}{2\sin\beta\cos\