Can I get a detailed explanation as of how this is solved, not just the math work?

#csc^2x tan^2x - 1 = tan^2x#

1 Answer
Feb 18, 2018

Sure, I'll write the steps right next to the math as I'm doing them.

First, we have to know the reciprocal functions. For this problem, the ones you will need to know are:

#cscx=1/sinx#

#secx=1/cosx#

You can also square both sides of each of these identities and get two more very similar ones:

#csc^2x=1/sin^2x#

#sec^2x=1/cos^2x#

Next, we have to know an identity. Using #sin^2x+cos^2x=1#, we can derive a new identity if we divide the whole equation by #cos^2x#:

#color(white)=>sin^2x+cos^2x=1#

#=>sin^2x/cos^2x+cos^2x/cos^2x=1/cos^2x#

#=>tan^2x+1=sec^2x#

If we rearrange the terms in this identity, we get the one that we need:

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

Now, here's the actual problem. We'll start with the left side, then manipulate it so that it will equal the right side:

#color(white){color(black)( (csc^2x*tan^2x-1, "Left side of the problem"), (1/sin^2x*sin^2x/cos^2x-1,"Write everything in terms of sin and cos"), ((1*sin^2x)/(sin^2x*cos^2x)-1, "Combine the fractions"), (sin^2x/(sin^2x*cos^2x)-1, "Simplify"), (color(red)cancel(color(black)(sin^2x))/(color(red)cancel(color(black)(sin^2x))*cos^2x)-1, "Cancel the like terms"), (1/cos^2x-1, "Rewrite"), (sec^2x-1, "Write the reciprocal function of"cos^2", which is"sec^2), (tan^2x, "Use the identity we proved before")):}#

That's it! We just proved that the left side of the equation equals the right side.

I hope this explanation was of help to you!