Answer
The proof is as shown below.
Work Step by Step
Consider the figure given below.
Now, we have $\tan\theta=c$ for the right angled triangle $ABC$.
By definition, $\tan\theta=\dfrac{\text{Perpendicular with respect to angle }\theta}{\text{Base with respect to the angle }\theta}=\dfrac{AB}{BC}$, so, we get
$\dfrac{AB}{BC}=c$, or, $AB=c\cdot BC$.
Now, using the Pythagorean Theorem in the triangle $ABC$:
$(AC)^2=(AB)^2+(AC)^2$
$\implies (AC)^2=(c\cdot BC)^2+(BC)^2$
$\implies (AC)^2=c^2 (BC)^2+(BC)^2$ [Using exponent rule]
$\implies (AC)^2=(c^2+1)(BC)^2$
$\implies \dfrac{(AC)^2}{(BC)^2}=c^2+1$
$\implies \dfrac{(AC)^2}{(BC)^2}=c^2+1$ [Taking square root on both sides]
$\implies \dfrac{AC}{BC}=\sqrt{c^2+1}$
$\implies \dfrac{BC}{AC}=\dfrac{1}{\sqrt{c^2+1}}$
By definition, $\cos\theta=\dfrac{\text{Base with respect to angle }\theta}{\text{Hypotenuse}}=\dfrac{BC}{AC}=\dfrac{1}{\sqrt{c^2+1}}$, hence, proved.
