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

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

Mathematics
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yes, first substitute \[4\tan \theta\] for x
so you have\[\sqrt{16+16\tan ^{2}\theta}\]
I am assuming that you would plug in 4tanθ to get: ` √(16+(4tanθ)^2) just not sure where to go from there

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Other answers:

you can factor out a 16 and take it out of the radical \[\sqrt{16(1+\tan ^{2}\theta)} = 4\sqrt{(1+\tan ^{2}\theta)}\]
now you need to make a trig substitution. \[1+\tan ^{2}\theta=\sec ^{2}\theta\]
OK... so when you say take it out of the radical you mean...? sorry this is where I get confused
since we had \[\sqrt{16(1+\tan ^{2}\theta)}\] we can evaluate the square root of 16
Ohhh ok
and take it out of the square root
so then the final step would be \[4\sqrt{\sec ^{2}\theta}=4\sec \theta\]
does that make sense?
sort of... I'm trying to figure out how you got 1+tan^2θ=sec^2θ to turn into 4√(sec^2θ)=4secθ?
ok
\[4\sqrt{1+\tan ^{2}\theta}=4\sqrt{\sec ^{2}\theta}\]
I first just replaced 1+tan^2θ with sec^2θ
then, we can take the square root of sec^2θ
which is just secθ
oh ok that makes sense thanks!
No problem
say I had one with cot.... √(9+(3cot^2θ) √(9+9cotθ)?
or I mean √(9+9cot^2θ)?
is that right
ok then I think....1√(1+cot^2θ) how do I know what to substitute then?
Or I mean 9√(1+cot^2θ)

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