Find an algebraic expression that is equivalent to the expression. $\sec ( \arctan 4 x )$
A) $\sqrt { 16 x ^ { 2 } + 1 }$
B) $\sqrt { 1 - 16 x ^ { 2 } }$
C) $\frac { \sqrt { 16 x ^ { 2 } - 1 } } { 16 x ^ { 2 } - 1 }$
D) $\sqrt { 16 x ^ { 2 } - 1 }$
E) $\frac { \sqrt { 16 x ^ { 2 } + 1 } } { 16 x ^ { 2 } + 1 }$

Question

Find an algebraic expression that is equivalent to the expression.   $\sec ( \arctan 4 x )$  
A) $\sqrt { 16 x ^ { 2 } + 1 }$
B) $\sqrt { 1 - 16 x ^ { 2 } }$
C) $\frac { \sqrt { 16 x ^ { 2 } - 1 } } { 16 x ^ { 2 } - 1 }$
D) $\sqrt { 16 x ^ { 2 } - 1 }$
E) $\frac { \sqrt { 16 x ^ { 2 } + 1 } } { 16 x ^ { 2 } + 1 }$

Quizplus · Accepted Answer

The expression $\sec(\arctan(4x))$ can be simplified using trigonometric identities. Consider a right triangle where the tangent of an angle is $4x$, meaning the opposite side over the adjacent side is $4x$. If we let the opposite side be $4x$ and the adjacent side be $1$, the hypotenuse, using the Pythagorean theorem, is $\sqrt{1^2 + (4x)^2} = \sqrt{16x^2 + 1}$. Since secant is the reciprocal of cosine, and cosine is adjacent over hypotenuse, $\sec(\arctan(4x))$ simplifies to the reciprocal of the cosine, which is the hypotenuse over the adjacent side, resulting in $\sqrt{16x^2 + 1}$.

Find an Algebraic Expression That Is Equivalent to the Expression sec⁡(arctan⁡4x)\sec ( \arctan 4 x )sec(arctan4x)

Find an Algebraic Expression That Is Equivalent to the Expression $\sec ( \arctan 4 x )$