## Trigonometry (11th Edition) Clone

$$\frac{1}{\sec\alpha-\tan\alpha}=\sec\alpha+\tan\alpha$$ The formula is proved to be an identity.
$$\frac{1}{\sec\alpha-\tan\alpha}=\sec\alpha+\tan\alpha$$ Let's transform the right side first, using $\sec\alpha=\frac{1}{\cos\alpha}$ and $\tan\alpha=\frac{\sin\alpha}{\cos\alpha}$ $$\sec\alpha+\tan\alpha$$ $$=\frac{1}{\cos\alpha}+\frac{\sin\alpha}{\cos\alpha}$$ $$=\frac{1+\sin\alpha}{\cos\alpha}$$ The right side looks like there is nothing more that should be done right now, so we move to transform the left side, using the same identities mentioned above. $$\frac{1}{\sec\alpha-\tan\alpha}$$ $$=\frac{1}{\frac{1}{\cos\alpha}-\frac{\sin\alpha}{\cos\alpha}}$$ $$=\frac{1}{\frac{1-\sin\alpha}{\cos\alpha}}$$ $$=\frac{\cos\alpha}{1-\sin\alpha}$$ Now both sides look almost the same, but not quite. We would try to make them the same by multiplying both numerator and denominator of the fraction on the left side with $(1+\sin\alpha)$ $$=\frac{\cos\alpha(1+\sin\alpha)}{(1-\sin\alpha)(1+\sin\alpha)}$$ $$=\frac{\cos\alpha(1+\sin\alpha)}{1-\sin^2\alpha}$$ $$=\frac{\cos\alpha(1+\sin\alpha)}{\cos^2\alpha}$$ (for $\cos^2\alpha=1-\sin^2\alpha$) $$=\frac{1+\sin\alpha}{\cos\alpha}$$ Now we see that both sides are equal, the formula is thus proved to be an identity.