The value of $$\frac{1}{\sin \theta} - \frac{\cot^2 \theta}{1 + \cosec \theta}$$ is:

$$\frac{1}{\sin \theta} - \frac{\cot^2 \theta}{1 + \cosec \theta}$$

where θ is replaced by t

Using cot(t)=$$\frac{cost}{sint}$$ and cosec=$$\frac{1}{sint}$$

$$\frac{1}{sint}$$ - $$\frac{cot^2t}{1+cosect}$$

$$\frac{1}{sint}$$ - $$\frac{cos^2t÷sin^2t}{1+(1÷sint)}$$

$$\frac{1}{sint}$$ - $$\frac{cos^2t÷sin^2t}{(sint+1)÷sint}$$

$$\frac{1}{sint}$$ - $$\frac{cos^2t}{sin^2t}$$ × $$\frac{sint}{sint+1}$$

$$\frac{1}{sint}$$ - $$\frac{cos^2t}{sint(sint+1)}$$

use this identity  $$\cos^2 \theta +\sin^2 \theta$$=1

hence $$\frac{1}{sint}$$ - $$\frac{1-sin^2t}{sint(sint+1)}$$

Use this identity $$(a^2-b^2)$$= (a+b)(a-b)

=$$\frac{1}{sint}$$ - $$\frac{(1+sint) × (1-sint )}{sint(sint+1)}$$

=$$\frac{1}{sint}$$ - $$\frac{1-sint }{sint}$$

=$$\frac{1-(1-sint)}{sint}$$

=$$\frac{1-(1+sint)}{sint}$$

=$$\frac{sint}{sint}$$

= 1