$a)L=\frac{{\sin }^{2}x-{\mathrm{tg}}^{2}x}{{\cos }^{2}x-{\mathrm{ctg}}^{2}x}=\frac{{\sin }^{2}x-\frac{{\sin }^{2}x}{{\cos }^{2}x}}{{\cos }^{2}x-\frac{{\cos }^{2}x}{{\sin }^{2}x}}=\frac{\frac{{\sin }^{2}x{\cos }^{2}x}{{\cos }^{2}x}-\frac{{\sin }^{2}x}{{\cos }^{2}x}}{\frac{{\sin }^{2}x{\cos }^{2}x}{{\sin }^{2}x}-\frac{{\cos }^{2}x}{{\sin }^{2}x}}=\frac{\frac{{\sin }^{2}x\left({\cos }^{2}x-1\right)}{{\cos }^{2}x}}{\frac{{\cos }^{2}x\left({\sin }^{2}x-1\right)}{{\sin }^{2}x}}=\frac{-\frac{{\sin }^{4}x}{{\cos }^{2}x}}{-\frac{{\cos }^{4}x}{{\sin }^{2}x}}=\frac{{\sin }^{4}x}{{\cos }^{2}x}\cdot \frac{{\sin }^{2}x}{{\cos }^{4}x}=\frac{{\sin }^{6}x}{{\cos }^{6}x}={\mathrm{tg}}^{6}x=P$