Simplify #(tan^2x+1)(cos^2x-1)#?

Use the Pythagorean Identities to make the both sides of the equation look identical, but only modify ONE SIDE.

The 3 Pythagorean Identities are as follows, but you will only use 1 & 2 for this problem.
1) #sin^2x+cos^2x=1#
2) #tan^2x+1=sec^2x#
3) #cot^2x+1=csc^2x#

NOTE: I chose to make the left side identical to the right side so that #-tan^2x=-tan^2x#, meaning I only worked with the left side.

#(tan^2x+1)(cos^2x+1)=-tan^2x#

STEP 1: SUBSTITUTE TERMS USING PYTHAGOREAN IDENTITIES.
1) Using the third identity, substitute #(sec^2x)# for #(tan^2x+1)#.
#(sec^2x)(cos^2x+1)=-tan^2x#

2) Notice that the first identity can be rewritten as #cos^2x-1=-sin^2x#. Substitute #(-sin^2x)# for #(cos^2x+1)#.*
#(sec^2x)(-sin^2x)=-tan^2x#

STEP 2: REWRITE TERMS USING ONLY SINE & COSINE TO HELP SIMPLIFY.
#(1/cos^2x)(-sin^2x)=-tan^2x#

STEP 3: SIMPLIFY SO THAT BOTH SIDES OF THE EQUATION ARE IDENTICAL.
#-sin^2x/cos^2x=-tan^2x#
#-tan^2x=-tan^2x#
QED

Finish off your little proof with either "QED" or a shaded box.

#(tan^2x+1)(cos^2x-1)#

= #sec^2x xx (-(1-cos^2x))#

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

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

= #-tan^2x#