Question

Prove secx-1/tanx= tanx/secx+1

Answer 1

Begin with the LHS:
\[\frac{\sec(x) - 1}{\tan(x)}\]

Multiply by \(\frac{\sec(x) + 1}{\sec(x) + 1}\):
\[\frac{\sec(x) - 1}{\tan(x)} \dot\ \frac{\sec(x) + 1}{\sec(x) + 1}\]

\[\frac{(\sec(x) - 1)(\sec(x) + 1)}{\tan(x)(\sec(x) + 1)}\]

Notice that \((\sec(x) - 1)(\sec(x) + 1) = \sec^2x - 1\) since \((a + b)(a - b) = a^2 - b^2\)
\[\frac{\sec^2x - 1}{\tan(x)(\sec(x) + 1)}\]

Remember that \(\sec^2x - 1 = \tan^2x\)

So
\[\frac{\tan^2x}{\tan(x)(\sec(x) + 1)}\]

Now simplify