How do you determine graphically and analytically whether #y_1=sec^2x-1# is equivalent to #y_2=tan^2x#?

1 Answer
Feb 24, 2018

Answer:

See below.

Explanation:

Identities:

1) #color(white)(88)color(red)bb(sin^2x+cos^2x=1)#

2)#color(white)(88)color(red)bb(sec^2x=1/cos^2x)#

3)#color(white)(88)color(red)bb(tan^2x=sin^2x/cos^2x)#

#y_1=y_2=>sec^2x-1=tan^2x#

LHS

By identity 2.

#1/cos^2x-1#

Add:

#(1-cos^2x)/cos^2x#

From identity 1:

#1-cos^2x=sin^2x#

Hence:

#sin^2x/cos^2x#

By identity 3:

#sin^2x/cos^2x=tan^2x#

So:

#LHS-=RHS#

This is an identity, so they are equivalent.

Graphically:

Plot both #bb(sec^2x-1)# and #bb(tan^2x)#. As you can see their graphs are identical, as we would expect.

#bb(y=sec^2x-1)#

graph{y=(tan(x))^2 [-11.25, 11.25, -5.625, 5.625]}

#bb(y=tan^2x)#
graph{y=(sec(x))^2-1 [-11.25, 11.25, -5.625, 5.625]}