$\frac{1}{1-\sin\left(x\right)}+\frac{-1}{1+\sin\left(x\right)}=2\tan\left(x\right)\sec\left(x\right)$

$\frac{1}{1-sin\left(x\right)}-\frac{1}{1+\sin\left(x\right)}=2\tan\left(x\right)sec\left(x\right)$

$\frac{1-sin\left(x\right)}{cos\left(x\right)}=\frac{cos\left(x\right)}{1+sin\left(x\right)}$

$\frac{1}{sec\left(x\right)+tan\left(x\right)}=sec\left(x\right)-tan\left(x\right)$

$\frac{1-\sin\:\left(x\right)}{\cos\:\left(x\right)}=\frac{\cos\:\left(x\right)}{1+\sin\:\left(x\right)}$

$\left(1+sinx\right)\left(1-sinx\right)=cos^2x$

$\sec\:\left(x\right)-\tan\:\left(x\right)=\frac{\cos\:\left(x\right)}{1+\sin\:\left(x\right)}$

$tan\left(x\right)+cot\left(x\right)=sec\left(x\right)cosec\left(x\right)$