If x = 4 cot θ,use trigonometric substitution to write $\sqrt { 16 + x ^ { 2 } }$ as a trigonometric function of θ,where 0

Question

If x = 4 cot θ,use trigonometric substitution to write $\sqrt { 16 + x ^ { 2 } }$ as a trigonometric function of θ,where 0 < θ < π.
A)​4 cos θ
B)​4 csc θ
C)​4 cot θ
D)​4 sec θ
E)​4 sin θ

Quizplus · Accepted Answer

If we substitute x = 4 cot θ, then cot θ = x/4. Using the Pythagorean identity, we get:

1 + cot2 θ = csc2 θ
1 + (x/4)2 = csc2 θ
csc θ = sqrt(16 + x2)/4

Substituting this back into the original expression, we get:

11ea99db_b4ab_15fe_94ac_c7e3e6077e08_TB4590_11 = 16/(sqrt(16 + x2)) = 16/[sqrt(16 + (4 cot θ)2)]
= 16/[sqrt(16 + 16 cot2 θ)]
= 16/[4 sqrt(1 + cot2 θ)]
= 4 csc θ

Therefore, the answer is B) 4 csc θ.

If X = 4 Cot θ,Use Trigonometric Substitution to Write 16+x2\sqrt { 16 + x ^ { 2 } }16+x2​

If X = 4 Cot θ,Use Trigonometric Substitution to Write $\sqrt { 16 + x ^ { 2 } }$