## anonymous 5 years ago If x=4tanθ, use trigonometric substitution to write √(16+x^2) as a trigonometric function of θ, where 0<θ<π/2

1. anonymous

yes, first substitute $4\tan \theta$ for x

2. anonymous

so you have$\sqrt{16+16\tan ^{2}\theta}$

3. anonymous

I am assuming that you would plug in 4tanθ to get: ` √(16+(4tanθ)^2) just not sure where to go from there

4. anonymous

you can factor out a 16 and take it out of the radical $\sqrt{16(1+\tan ^{2}\theta)} = 4\sqrt{(1+\tan ^{2}\theta)}$

5. anonymous

now you need to make a trig substitution. $1+\tan ^{2}\theta=\sec ^{2}\theta$

6. anonymous

OK... so when you say take it out of the radical you mean...? sorry this is where I get confused

7. anonymous

since we had $\sqrt{16(1+\tan ^{2}\theta)}$ we can evaluate the square root of 16

8. anonymous

Ohhh ok

9. anonymous

and take it out of the square root

10. anonymous

so then the final step would be $4\sqrt{\sec ^{2}\theta}=4\sec \theta$

11. anonymous

does that make sense?

12. anonymous

sort of... I'm trying to figure out how you got 1+tan^2θ=sec^2θ to turn into 4√(sec^2θ)=4secθ?

13. anonymous

ok

14. anonymous

$4\sqrt{1+\tan ^{2}\theta}=4\sqrt{\sec ^{2}\theta}$

15. anonymous

I first just replaced 1+tan^2θ with sec^2θ

16. anonymous

then, we can take the square root of sec^2θ

17. anonymous

which is just secθ

18. anonymous

oh ok that makes sense thanks!

19. anonymous

No problem

20. anonymous

say I had one with cot.... √(9+(3cot^2θ) √(9+9cotθ)?

21. anonymous

or I mean √(9+9cot^2θ)?

22. anonymous

is that right

23. anonymous

ok then I think....1√(1+cot^2θ) how do I know what to substitute then?

24. anonymous

Or I mean 9√(1+cot^2θ)