Question

Prove the following identity: (sec a + tan a)/(sec a - tan a) = (sec a + tan a)^2

Answer 1

can we meet in the middle?

Answer 2

yes

Answer 3

the right hand side is
\[\frac{\sec(x)+\tan(x)}{\sec(x)-\tan(x)}=\frac{\frac{1}{\cos(x)}+\frac{\sin(x)}{\cos(x)}}{\frac{1}{\cos(x)}-\frac{\sin(x)}{\cos(x)}}\]

Answer 4

multiply top and bottom by cosine to get
\[\frac{1+\sin(x)}{1-\sin(x)}\]

Answer 5

sorry that was the left hand side, not the right hand side

Answer 6

the right hand side is
\[(\frac{1}{\cos(x)}+\frac{\sin(x)}{\cos(x)})^2\]
\[\frac{1+2\sin(x)+\sin^2(x)}{\cos^2(x)}=\frac{(1+\sin(x))^2}{\cos^2(x)}\]

Answer 7

so far no trig at all, just algebra.

Answer 8

now write the denominator of the right hand side as
\[1-\sin^2(x)=(1+\sin(x))(1-\sin(x))\] and cancel with one of the factors in the numerator. you get
\[\frac{1+\sin(x)}{1-\sin(x)}\] which is what you were looking for