LHS: $(\cosec \theta - \cot \theta)^2$

We know $\cosec \theta = \frac{1}{\sin \theta}$ and $\cot \theta = \frac{\cos \theta}{\sin \theta}$.

So, LHS

$$= \left(\frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta}\right)^2 = \left(\frac{1 - \cos \theta}{\sin \theta}\right)^2 = \frac{(1 - \cos \theta)^2}{\sin^2 \theta}$$

Using identity $\sin^2 \theta = 1 - \cos^2 \theta = (1 - \cos \theta)(1 + \cos \theta)$,

$$\text{LHS} = \frac{(1 - \cos \theta)^2}{(1 - \cos \theta)(1 + \cos \theta)} = \frac{1 - \cos \theta}{1 + \cos \theta} = \textbf{RHS}$$

Hence proved.

LHS: $\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A}$

Taking common denominator $\cos A (1 + \sin A)$,

$$= \frac{\cos^2 A + (1 + \sin A)^2}{\cos A (1 + \sin A)}$$

Expand numerator:

$$\cos^2 A + 1 + 2\sin A + \sin^2 A = (\cos^2 A + \sin^2 A) + 1 + 2\sin A = 1 + 1 + 2\sin A = 2(1 + \sin A)$$

Thus LHS

$$= \frac{2(1 + \sin A)}{\cos A (1 + \sin A)} = \frac{2}{\cos A} = 2 \sec A = \textbf{RHS}$$

Hence proved.

LHS: $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta}$

Express in terms of $\sin \theta$ and $\cos \theta$: $\tan \theta = \frac{\sin \theta}{\cos \theta}$, $\cot \theta = \frac{\cos \theta}{\sin \theta}$.

First term:

$$\frac{\sin \theta/\cos \theta}{1 - \cos \theta/\sin \theta} = \frac{\sin \theta/\cos \theta}{(\sin \theta - \cos \theta)/\sin \theta} = \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)}$$

Second term:

$$\frac{\cos \theta/\sin \theta}{1 - \sin \theta/\cos \theta} = \frac{\cos \theta/\sin \theta}{(\cos \theta - \sin \theta)/\cos \theta} = \frac{\cos^2 \theta}{\sin \theta (\cos \theta - \sin \theta)} = -\frac{\cos^2 \theta}{\sin \theta (\sin \theta - \cos \theta)}$$

Adding:

$$\text{LHS} = \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta (\sin \theta - \cos \theta)} = \frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$$

Factor numerator:

$$\sin^3 \theta - \cos^3 \theta = (\sin \theta - \cos \theta)(\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta)$$

Cancel $(\sin \theta - \cos \theta)$:

$$\text{LHS} = \frac{\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} + 1 = \cosec \theta \sec \theta + 1 = 1 + \sec \theta \cosec \theta = \textbf{RHS}$$

Hence proved.

LHS:

$$\frac{1 + \sec A}{\sec A} = \frac{1}{\sec A} + \frac{\sec A}{\sec A} = \cos A + 1$$

RHS: $\frac{\sin^2 A}{1 - \cos A}$

Using identity $\sin^2 A = 1 - \cos^2 A = (1 - \cos A)(1 + \cos A)$,

$$\text{RHS} = \frac{(1 - \cos A)(1 + \cos A)}{1 - \cos A} = 1 + \cos A$$

Thus LHS $=$ RHS. Hence proved.

LHS: $\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1}$

Divide numerator and denominator by $\sin A$ ($\sin A \neq 0$ for acute $A$):

$$= \frac{\cot A - 1 + \cosec A}{\cot A + 1 - \cosec A}$$

Let $x = \cot A$, $y = \cosec A$. Then $y^2 - x^2 = 1$ (using $\cosec^2 A = 1 + \cot^2 A$).

Consider

$$(x + y)(x - y + 1) = (x+y)(x-y) + (x+y) = (x^2 - y^2) + (x+y) = -1 + (x+y)$$

Thus $(x + y)(x - y + 1) = x + y - 1$, which gives

$$\frac{x + y - 1}{x - y + 1} = x + y$$

Substituting back, LHS

$$= x + y = \cot A + \cosec A = \cosec A + \cot A = \textbf{RHS}$$

Hence proved.

LHS: $\sqrt{\frac{1 + \sin A}{1 - \sin A}}$

Multiply numerator and denominator inside the square root by $(1 + \sin A)$:

$$= \sqrt{\frac{(1 + \sin A)^2}{1 - \sin^2 A}}$$

Using $1 - \sin^2 A = \cos^2 A$, we get

$$= \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}} = \frac{1 + \sin A}{|\cos A|}$$

Since $A$ is acute, $\cos A > 0$, so

$$= \frac{1 + \sin A}{\cos A} = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \sec A + \tan A = \textbf{RHS}$$

Hence proved.

LHS: $\frac{\sin \theta - 2\sin^3 \theta}{2\cos^3 \theta - \cos \theta}$

Factor $\sin \theta$ from numerator and $\cos \theta$ from denominator:

$$= \frac{\sin \theta (1 - 2\sin^2 \theta)}{\cos \theta (2\cos^2 \theta - 1)}$$

Using $\sin^2 \theta + \cos^2 \theta = 1$, we have $1 - 2\sin^2 \theta = \cos^2 \theta - \sin^2 \theta$ and $2\cos^2 \theta - 1 = \cos^2 \theta - \sin^2 \theta$.

Thus numerator $= \sin \theta (\cos^2 \theta - \sin^2 \theta)$, denominator $= \cos \theta (\cos^2 \theta - \sin^2 \theta)$.

Cancel $(\cos^2 \theta - \sin^2 \theta)$ (non-zero for acute $\theta$):

$$\text{LHS} = \frac{\sin \theta}{\cos \theta} = \tan \theta = \textbf{RHS}$$

Hence proved.

LHS: $(\sin A + \cosec A)^2 + (\cos A + \sec A)^2$

Expand:

$$= (\sin^2 A + 2\sin A \cosec A + \cosec^2 A) + (\cos^2 A + 2\cos A \sec A + \sec^2 A)$$

Since $\sin A \cosec A = 1$ and $\cos A \sec A = 1$, we get

$$= \sin^2 A + \cos^2 A + 2 + 2 + \cosec^2 A + \sec^2 A = 1 + 4 + \cosec^2 A + \sec^2 A = 5 + \cosec^2 A + \sec^2 A$$

Using identities $\cosec^2 A = 1 + \cot^2 A$ and $\sec^2 A = 1 + \tan^2 A$, we obtain

$$\text{LHS} = 5 + (1 + \cot^2 A) + (1 + \tan^2 A) = 7 + \tan^2 A + \cot^2 A = \textbf{RHS}$$

Hence proved.