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

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Similar Problems Solved

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

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

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

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

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

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

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

Main Topic: Identità trigonometriche

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