Can someone help me prove that 3sec(x)^2-2=sec(x)^2+2tan(x)^2 ?

[abbb7615]

I can't find a way to prove that

 

[mmm4444bot]

I can't find a way to prove that

Use identities for secant and tangent to rewrite each side in terms of cosine and sine.

Multiply each side by cos(θ)^2.

Does the result give you any ideas? (Hint: There's another identity waiting to be used.) :cool:

 

[abbb7615]

Thanks for your answer. I found an easier solution.

3sec(x)² - 2 = sec(x)² + 2tan(x)²
3sec(x)² = sec(x)² + 2( 1 + tan(x)² )
3sec(x)² = sec(x)² + 2sec(x)²

What do you think of this solution ?

 

[Deleted member 4993]

Thanks for your answer. I found an easier solution.

3sec(x)² - 2 = sec(x)² + 2tan(x)²
3sec(x)² = sec(x)² + 2( 1 + tan(x)² )
3sec(x)² = sec(x)² + 2sec(x)²

What do you think of this solution ?

In my opinion, a cleaner version would be:

3sec²(x) - 2

= sec²(x) + 2sec²(x) - 2

= sec²(x) + 2[sec²(x) - 1]

= sec²(x) + 2[tan²(x)]

 

[mmm4444bot]

3sec(x)² - 2 = sec(x)² + 2tan(x)²
3sec(x)² = sec(x)² + 2( 1 + tan(x)² )
3sec(x)² = sec(x)² + 2sec(x)²

What do you think of this solution ?

I think it's great.

All three methods in this thread (so far) are solid. :cool:

 

[abbb7615]

Thank you everybody. I'm glad I found this great forum.