If Tan^2(t)-Sin^2(t)=Sin^a(t)/Cos^b(t), then the positive power a= ALSO the positive power b=

Question

So, I know the first two steps

sin^2(t)/cos^2(t) -sin^2(t)

sin^2(t)-sin^2(t)cos^2(t)/(cos^2(t)

I have no idea what to do after this? I looked and it said to factor but I have no idea why you factor sin^2(t)???

^This is me just simplifying the left side.

Please show all steps so I can understand how to get the answer! Thank you!

Answer 1

First, we know that tan t = (sin t) / (cos t), so

sin² t

tan² t = _______

cos² t

Rewriting the original questiona tan² t - sin² t is the same as

sin² t

= _______ - sin² t

cos² t

We want to express this as a single fraction. Right now, we have the fractions,

sin² t sin² t

_______ - ________

cos² t 1

In order to combine the fractions, we need a comm on denominator. Our common denominator would be cos² t. But if we only change the denominator in the second fraction from 1 to cos² t, we've changed the problem. The only value you can multiply the fraction by and not make a new problem is 1. But we still want the denominator to equal cos² t. To do this, we can multiply the fraction by cos² t / cos² t since anything divided by itself is 1.

sin² t cos² t sin² t * cos² t

________ * _________ = ________________

1 cos² t cos² t

sin² t sin² t * cos² t

= ________ - _______________

cos² t cos² t

sin² t - (sin² t * cos² t)

= __________________________

cos²t

Both terms have sin² t in common, so we can factor it out,

sin² t * ( 1 - cos² t)

= ______________________

cos² t

The Pythagorean Identity tells us that cos² t + sin² t = 1. If we subtract cos² t from both sides, this gives us: sin² t = 1 - cos² t

Therefore, we can substitute our ( 1 - cos² t) with sin² t in the numerator, giving us

sin² t * sin² t

= __________________

cos² t

= sin⁴ t

_____________

cos² t

Therefore, a = 4 and b = 2