Verify sec^2x-tan^2x=tanxcotx by changing only one side?

To solve this, we will have to use the Reciprocal Identity and the Pythagorean Identity.

The Reciprocal Identity basically defines #csc#, #sec#, #cot# from the known #sin#, #cos#, and #tan#, respectively.

The Pythagorean Identity is defined as:

#cos^2 + sin^2 = 1#

We can manipulate this into two parts:

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

These will come in handy here in a moment.

Knowing these, all we really have to do is change the left side accordingly:

#(1/cos)^2 x - (sin/cos)^2 x = tanx cotx#

This can be further simplified into:

#((1-sin^2)/cos^2)x = tanx cotx#

#(cos^2/cos^2)x = tanx cotx#

#1 = tanx cotx#

From here, we aren't necessarily changing the right side, but we are simplifying it.

Since #tanx = (sin/cos)# and #cotx = (cos/sin)# they would both cancel out and become one.

Therefore, #sec^2x - tan^2x = tanx cotx# would validate to 1 = 1.

.

#sec^2x-tan^2x=tanxcotx#

#sec^2x-tan^2x=1/cos^2x-sin^2x/cos^2x=(1-sin^2x)/cos^2x=cos^2x/cos^2x=cos^2x/cos^2x*(sinxcosx)/(sinxcosx)=cosx/cosx*cosx/cosx*sinx/cosx*cosx/sinx=(cosx/cosx*sinx/cosx)(cosx/cosx*cosx/sinx)=(cancelcolor(red)cosx/cosx*sinx/cancelcolor(red)cosx)(cosx/cancelcolor(red)cosx*cancelcolor(red)cosx/sinx)=(sinx/cosx)(cosx/sinx)=tanxcotx#