Odpowiedź :
[tex](1+\sin\alpha)\left(\frac{1}{\cos\alpha}-\frac{1}{\text{ctg}}\right)=\cos\alpha[/tex]
Założenie:
[tex]\cos\alpha\neq 0\land \text{ctg}\alpha\neq 0\\\cos\alpha\neq 0\land\frac{\cos\alpha}{\sin\alpha}\neq 0\\\cos\alpha\neq 0\land\sin\alpha\neq 0\\\alpha\neq \frac{\pi}{2}+k\pi\land \alpha\neq k\pi\land k\in\mathbb{Z}\\\alpha\neq \frac{k\pi}{2}\land k\in\mathbb{Z}[/tex]
Sprawdzenie tożsamości:
[tex](1+\sin\alpha)\left(\frac{1}{\cos\alpha}-\frac{1}{\text{ctg}}\right)=(1+\sin\alpha)\left(\frac{1}{\cos\alpha}-\frac{1}{\frac{\cos\alpha}{\sin\alpha}}\right)=(1+\sin\alpha)\left(\frac{1}{\cos\alpha}-\frac{\sin\alpha}{\cos\alpha}\right)=\\=(1+\sin\alpha)\frac{1-\sin\alpha}{\cos\alpha}=\frac{(1+\sin\alpha)(1-\sin\alpha)}{\cos\alpha}=\frac{1-\sin^2\alpha}{\cos\alpha}=\frac{\cos^2\alpha}{\cos\alpha}=\cos\alpha[/tex]