Question

Answer this question ​

Answer 1

Answer:

D.

$\frac{b}{ \sqrt{{a}^{2} + {b}^{2} } }$

Step-by-step explanation:

One of the angles measure

${tan}^{ - 1} ( \frac{a}{b} )$

So,

$\tan(θ) = \frac{a}{b}$

With this it is clear that a is perpendicular and b is base as tanθ=Perpendicular/Base

and we already know that

$\sqrt{ {a}^{2} + {b}^{2} }$

is hypotenuse as it is the side opposite to perpendicular angle.

Now, To find

$cos( {tan}^{ - 1} ( \frac{a}{b} ))$

we already know that

${tan}^{ - 1} ( \frac{a}{b} ) = θ$

then,

$cos( {tan}^{ - 1} ( \frac{a}{b} )) = \cos(θ)$

and we know that cosθ= Base / Hypotenuse

then

$\cos(θ) = \frac{b}{ \sqrt{ {a}^{2} + {b}^{2} } }$

Therefore,

$cos( {tan}^{ - 1} ( \frac{a}{b} )) = \frac{b}{ \sqrt{ {a}^{2} + {b}^{2} } }$