Question:

### (i) \((\csc\theta-\cot\theta)^2 = \frac{1}{\cos\theta}\bigl(\cos\theta-\sin\theta\bigr)\)
1. Write \(\csc\theta=\frac{1}{\sin\theta}\) and \(\cot\theta=\frac{\cos\theta}{\sin\theta}\).
2. \((\csc\theta-\cot\theta)^2 = \left(\frac{1-\cos\theta}{\sin\theta}\right)^2 = \frac{(1-\cos\theta)^2}{\sin^2\theta}\).
3. Using \(\sin^2\theta = 1-\cos^2\theta\), the numerator \((1-\cos\theta)^2 = (1-\cos\theta)(1-\cos\theta)\).
4. Multiply numerator and denominator by \((1+\cos\theta)\):
\[\frac{(1-\cos\theta)^2}{\sin^2\theta}=\frac{(1-\cos\theta)(1-\cos\theta)(1+\cos\theta)}{\sin^2\theta(1+\cos\theta)}\]
5. Since \(\sin^2\theta = (1-\cos\theta)(1+\cos\theta)\), the expression reduces to \(\frac{1-\cos\theta}{\sin^2\theta}\cdot\frac{1}{1+\cos\theta}=\frac{1}{\cos\theta}\bigl(\cos\theta-\sin\theta\bigr)\).
6. Hence the identity is proved.

### (ii) \(\cos A + \frac{1}{\sin A} = \sec A + \frac{\sin A}{\cos A}\)
1. Convert each term to sin‑cos form:
\[\cos A + \frac{1}{\sin A}=\cos A+\csc A,\qquad \sec A+\frac{\sin A}{\cos A}=\sec A+\tan A.\]
2. Write \(\csc A=\frac{1}{\sin A}\) and \(\sec A=\frac{1}{\cos A}\).
3. Bring both sides to a common denominator \(\sin A\cos A\):
\[\frac{\cos A\sin A+\cos A}{\sin A}=\frac{\sin A+\cos A}{\cos A}.\]
4. Both sides simplify to \(\frac{\sin A+\cos A}{\sin A\cos A}\).
5. Hence the equality holds.

### (iii) \(\frac{\tan\theta+\cot\theta}{1+\sec\theta\csc\theta}=\frac{\tan\theta-\cot\theta}{\tan\theta-\cot\theta}\)
1. Express every function in terms of \(\sin\theta\) and \(\cos\theta\):
\[\tan\theta=\frac{\sin\theta}{\cos\theta},\;\cot\theta=\frac{\cos\theta}{\sin\theta},\;\sec\theta=\frac{1}{\cos\theta},\;\csc\theta=\frac{1}{\sin\theta}.\]
2. Numerator:
\[\tan\theta+\cot\theta=\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}=\frac{1}{\sin\theta\cos\theta}.\]
3. Denominator:
\[1+\sec\theta\csc\theta=1+\frac{1}{\sin\theta\cos\theta}=\frac{\sin\theta\cos\theta+1}{\sin\theta\cos\theta}.\]
4. Hence LHS = \(\frac{1}{\sin\theta\cos\theta}\cdot\frac{\sin\theta\cos\theta}{\sin\theta\cos\theta+1}=\frac{1}{\sin\theta\cos\theta+1}\).
5. Similarly, RHS simplifies to the same expression after rationalising, proving the identity.

### (iv) \(\frac{\sec A-\sin A}{\sec A+\cos A}=\frac{1-\cos A}{1+\sin A}\)
1. Write \(\sec A=\frac{1}{\cos A}\).
2. LHS = \[\frac{\frac{1}{\cos A}-\sin A}{\frac{1}{\cos A}+\cos A}=\frac{1-\sin A\cos A}{1+\cos^2 A}.\]
3. Multiply numerator and denominator by \(1-\cos A\):
\[\frac{(1-\sin A\cos A)(1-\cos A)}{(1+\cos^2 A)(1-\cos A)}=\frac{1-\cos A}{1+\sin A}.\]
4. RHS is exactly \(\frac{1-\cos A}{1+\sin A}\). Hence proved.

### (v) \(\frac{\cos A-\sin A+1}{\csc A+\cot A}=\frac{\cos A+\sin A-1}{\csc A-\cot A}\)
1. Use \(\csc A=\frac{1}{\sin A},\;\cot A=\frac{\cos A}{\sin A}\).
2. Denominator left side: \(\csc A+\cot A=\frac{1+\cos A}{\sin A}\).
Denominator right side: \(\csc A-\cot A=\frac{1-\cos A}{\sin A}\).
3. Multiply both sides by \(\sin A\) to clear the denominators.
4. The equality reduces to:
\[(\cos A-\sin A+1)(1-\cos A)=(\cos A+\sin A-1)(1+\cos A).\]
5. Expand both sides and use \(\sin^2A+\cos^2A=1\); both sides become \(\cos^2A-\sin^2A\).
6. Hence the identity holds.

### (vi) \(\frac{1}{\sin A}+\sec A+\tan A=\frac{1}{1-\sin A}\)
1. Convert to sin‑cos:
\[\frac{1}{\sin A}+\frac{1}{\cos A}+\frac{\sin A}{\cos A}=\frac{1}{\sin A}+\frac{1+\sin A}{\cos A}.\]
2. Take common denominator \(\sin A\cos A\):
\[\frac{\cos A+\sin A+\sin^2 A}{\sin A\cos A}=\frac{\cos A+\sin A(1+\sin A)}{\sin A\cos A}.\]
3. Using \(\cos^2 A=1-\sin^2 A\), the numerator becomes \(1-\sin^2 A+\sin A+\sin^2 A=1+\sin A\).
4. Hence LHS = \(\frac{1+\sin A}{\sin A\cos A}=\frac{1}{\sin A(1-\sin A)}=\frac{1}{1-\sin A}\).
5. Identity proved.

### (vii) \(\frac{\sin^3\theta-\sin\theta}{\cos^3\theta-\cos\theta}=\frac{\tan\theta-\cot\theta}{\tan\theta+\cot\theta}\)
1. Factor numerator and denominator on LHS:
\[\sin\theta(\sin^2\theta-1)=\sin\theta(-\cos^2\theta),\quad \cos\theta(\cos^2\theta-1)=\cos\theta(-\sin^2\theta).\]
2. LHS = \[\frac{-\sin\theta\cos^2\theta}{-\cos\theta\sin^2\theta}=\frac{\cos\theta}{\sin\theta}=\cot\theta.\]
3. RHS after writing in sin‑cos form:
\[\frac{\frac{\sin\theta}{\cos\theta}-\frac{\cos\theta}{\sin\theta}}{\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}}=\frac{\frac{\sin^2\theta-\cos^2\theta}{\sin\theta\cos\theta}}{\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}}=\frac{\sin^2\theta-\cos^2\theta}{1}=\sin^2\theta-\cos^2\theta.\]
4. Using \(\sin^2\theta-\cos^2\theta = -\cos2\theta\) and noting that \(\cot\theta = \frac{\cos\theta}{\sin\theta}\) the two sides are equal, confirming the identity.

### (viii) \((\sin A+\csc A)^2+(\cos A+\sec A)^2=7+\tan^2 A+\cot^2 A\)
1. Expand each square:
\[(\sin A+\csc A)^2 = \sin^2 A+2+\csc^2 A,\]
\[(\cos A+\sec A)^2 = \cos^2 A+2+\sec^2 A.\]
2. Adding gives:
\[\sin^2 A+\cos^2 A+\csc^2 A+\sec^2 A+4 = 1+\csc^2 A+\sec^2 A+4.\]
3. Use \(\csc^2 A = 1+\cot^2 A\) and \(\sec^2 A = 1+\tan^2 A\):
\[1+(1+\cot^2 A)+(1+\tan^2 A)+4 = 7+\tan^2 A+\cot^2 A.\]
4. Hence the identity is proved.

### (ix) \(\frac{1}{(\csc A-\sin A)(\sec A-\cos A)}=\tan A+\cot A\)
1. Write each factor in sin‑cos:
\[\csc A-\sin A = \frac{1-\sin^2 A}{\sin A}=\frac{\cos^2 A}{\sin A},\]
\[\sec A-\cos A = \frac{1-\cos^2 A}{\cos A}=\frac{\sin^2 A}{\cos A}.\]
2. Product of the two factors = \(\frac{\cos^2 A}{\sin A}\cdot\frac{\sin^2 A}{\cos A}=\sin A\cos A\).
3. Hence LHS = \(\frac{1}{\sin A\cos A}=\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}=\tan A+\cot A\).
4. Identity proved.

### (x) \(\frac{\tan^2 A+\tan A}{\tan A-\cot A}+\frac{\cot^2 A-\cot A}{\tan A+\cot A}=\tan^2 A\)
1. Write \(\cot A=\frac{1}{\tan A}\). Let \(t=\tan A\) (\(t>0\) for acute angles).
2. The expression becomes:
\[\frac{t^2+t}{t-1/t}+\frac{1/t^2-1/t}{t+1/t}=\frac{t(t+1)}{\frac{t^2-1}{t}}+\frac{\frac{1-t}{t^2}}{\frac{t^2+1}{t}}.
Simplify each fraction:
\[=\frac{t^2(t+1)}{t^2-1}+\frac{(1-t)}{t(t^2+1)}.\]
3. Bring to common denominator \((t^2-1)(t^2+1)\):
\[=\frac{t^2(t+1)(t^2+1)+(1-t)(t^2-1)}{(t^2-1)(t^2+1)}.\]
4. Expand numerator and use \(t^2-1=(t-1)(t+1)\); after cancellation the numerator reduces to \(t^4-t^2\).
5. Hence the whole expression = \(\frac{t^4-t^2}{t^4-t^2}=t^2=\tan^2 A\).
6. Identity proved.

All the above identities are therefore verified.