Question

PROVE: 1/secx-1 + 1/secx+1 = 2/cos tan^2x

Answer 1

well you need a common denominator... so cross multiply the for the numerators and get a common denominator

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

now the numerator simplifies

and the denominator is the difference of 2 squares

\[\frac{2 \sec(x)}{\sec^2(x) - 1}\]

now all yo need to do is realise

sec = 1/cos

and

sec^2 - 1 = tan^2

substitute them and you'll find your answer

Answer 2

You're supposed to solve for one of them only to get to the other....so I don't understand what you just did...

Answer 3

well I simplified the left hand side, basically I made it possible to add the 2 fractions... now you need to make some substitutions to the left hand side

Answer 4

oh...

Answer 5

i get it...thanks