Writing an Expression...2

[nycmathguy]

Section 4.7

Question 66

sec (arctan 3x)

Let u = arctan 3x

This means that tan u = 3x.

I know that tan u = opp/adj.

Using the Pythagorean Theorem, I found the hypotenuse to be sqrt{1 - 9x^2}.

After the u-substitution, the given problem becomes sec u.

I know that sec u = hyp/adj.

My answer is sec u = sqrt{1 - 9x^2}/1, which is simply sqrt{1 - 9x^2}.

However, the correct answer according to Photomath is sqrt{1 + 9x^2}.

Why is my answer wrong?

This ends Section 4.7.

 

[HallsofIvy]

Section 4.7

View attachment 1126

Question 66

sec (arctan 3x)

Let u = arctan 3x

This means that tan u = 3x.

I know that tan u = opp/adj.

Using the Pythagorean Theorem, I found the hypotenuse to be sqrt{1 - 9x^2}.

No, it's sqrt(1+ 9x^2).

After the u-substitution, the given problem becomes sec u.

I know that sec u = hyp/adj.

My answer is sec u = sqrt{1 - 9x^2}/1, which is simply sqrt{1 - 9x^2}.

However, the correct answer according to Photomath is sqrt{1 + 9x^2}.

Why is my answer wrong?

This ends Section 4.7.

 

[nycmathguy]

No, it's sqrt(1+ 9x^2).

I did try as you can see.