If $$\sec \theta - \tan \theta = P$$, then $$\cosec \theta = ?$$

As per the given question,

$$\Rightarrow \sec \theta - \tan \theta = P$$

$$\Rightarrow \dfrac{1}{\cos \theta} - \dfrac{\sin \theta}{\cos\theta} = P$$

$$\Rightarrow \dfrac{1-\sin \theta}{\cos \theta} = P$$

Squaring both side of the given equation,

$$\Rightarrow \dfrac{(1-\sin \theta)^2}{(\cos \theta)^2} = P^2$$

$$\Rightarrow \dfrac{(1-\sin \theta)^2}{1-\sin^2\theta} = P^2$$

$$\Rightarrow \dfrac{(1-\sin \theta)^2}{(1-\sin\theta)(1+\sin\theta)} = P^2$$

$$\Rightarrow \dfrac{(1-\sin \theta)}{(1+\sin\theta)} = P^2$$

$$\Rightarrow 1-\sin \theta = P^2(1+\sin\theta)$$

$$\Rightarrow 1-\sin \theta = P^2+P^2\sin\theta$$

$$\Rightarrow \sin\theta(P^2+1)=1-P^2$$

$$\Rightarrow \sin\theta=\dfrac{1-P^2}{(P^2+1)}$$

Hence $$\cosec \theta=\dfrac{P^2+1}{1-P^2}$$