tan^2x+sec^2x=1 true or false? for all values of x

Question

tan^2x+sec^2x=1 true or false?  for all values of x

Uhm I think false? 

 

 

tan^2 x+1=sec^2x.  So to get 1 on the other side of the equal sign wouldn't it be sec^2x-tan^2x=1?

 

I'm very confused... or 

 

could you change it into sin^2x/ cos^2x + 1/cos^2x=1  which would be sin^2x/cos^2x=1  which would be tan^2x=1

 

 

SO confused!  True or false?  Why! and how??  Thanks!!! :D

Answer 1

Cece,

 

I think you're just overthinking it.  Here's the statement that we're evaluating, at least as I'm reading the question:

 

"tan²x + sec²x = 1 is true for all values of x."

 

The identity, as you noted, is tan²x + 1 = sec²x, for all values of x.  Rearranging, you absolutely get:

 

tan²x - sec²x = 1.

 

So, the original statement is false.  Sure, there might be values of x for which the original equation works.  It's solvable, but that doesn't make it true for all x.

 

When you started messing with the equation by rewriting it as sines and cosines, I think you goofed the math:

 

sin²x/cos²x + 1/cos²x = 1.

 

After that, you said you'd get sin²x/cos²x = 1, which isn't true.  Ya lost the second term there, the 1/cos²x. If you multiply the whole thing by cos²x, you'd get:

 

sin²x + 1 = cos²x.

 

This brings us back to sines and cosines, and it's still not true for all x.  Therefore, it's still false.  :)

 

So yes, when you originally compared this to the identity you know for secants and tangents, that was how I was thinking about it, too.  It's not the same, so it's not true for all x.

 

 

Answer 2

It's false and true!

 

Remember this identity?

(sin(x))² + (cos(x))² = 1

 

Well, if we divide (cos(x))² on both sides; we get

 

(tan(x))² + 1 = (sec(x))²

1 = (sec(x))² - (tan(x))²

 

Now, we will see if 1 = (sec(x))² - (tan(x))² and 1 = (sec(x))² + (tan(x))² can both be true. We can do this by assuming that they are both true, and then add the equations to get

2 = 2(sec(x))²

1=(sec(x))²

 

But we know what values of x can make this true, don't we :)

 

Tutors, please feel free to review my explanation.