If ${\cos ^2}\theta - {\sin ^2}\theta = {\tan ^2}\phi $ then prove that $\cos \phi = \dfrac{1}{{\sqrt 2 \cos \theta }}$.

Hint:
We know that ${\cos ^2}\theta + {\sin ^2}\theta = 1$ so we can convert ${\sin ^2}\theta {\text{ into }}{\cos ^2}\theta $ and also we know that $\tan \phi = \dfrac{{\sin \phi }}{{\cos \phi }}$, So changing into $\cos \phi $, we will get our answer.

Complete step by step solution:
Here we are given the relation between the two angles $\theta {\text{ and }}\phi $ which are given as
${\cos ^2}\theta - {\sin ^2}\theta = {\tan ^2}\phi $ and we need to prove that $\cos \phi = \dfrac{1}{{\sqrt 2 \cos \theta }}$
Now as we know that $\sin \theta ,\cos \theta ,\tan \theta $ are all the trigonometric functions then $\sin \theta $ is the ratio of the length of the perpendicular and the hypotenuse in the right angles triangle whereas $\cos \theta $ is the ratio of the base and the hypotenuse in the right angle triangle.
And we should know that $\tan \theta $ is the ratio of the two trigonometric identities which are $\sin \theta \;and\cos \theta $
So we can say that $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ and also we know that ${\cos ^2}\theta + {\sin ^2}\theta = 1$ and here we are given that ${\cos ^2}\theta - {\sin ^2}\theta = {\tan ^2}\phi $ where $\theta {\text{ and }}\phi $ are two different angles and the relation between them is given.
Also we know that ${\cos ^2}\theta + {\sin ^2}\theta = 1$
So we can write it as ${\sin ^2}\theta = 1 - {\cos ^2}\theta $
Now putting the value of ${\sin ^2}\theta = 1 - {\cos ^2}\theta $ in the given equation we will get that
$\Rightarrow {\cos ^2}\theta - (1 - {\cos ^2}\theta ) = {\tan ^2}\phi $
$\Rightarrow {\cos ^2}\theta - 1 + {\cos ^2}\theta = {\tan ^2}\phi $
$\Rightarrow 2{\cos ^2}\theta - 1 = {\tan ^2}\phi $$ - - - - - (1)$
Now as we know that $\tan \phi = \dfrac{{\sin \phi }}{{\cos \phi }}$
So squaring both the sides we will get that
${\tan ^2}\phi = \dfrac{{{{\sin }^2}\phi }}{{{{\cos }^2}\phi }}$
So putting this value in equation (1) we will get that
$\Rightarrow 2{\cos ^2}\theta - 1 = \dfrac{{{{\sin }^2}\phi }}{{{{\cos }^2}\phi }}$
$\Rightarrow 2{\cos ^2}\theta - 1 = \dfrac{{1 - {{\cos }^2}\phi }}{{{{\cos }^2}\phi }}$
$
\Rightarrow 2{\cos ^2}\theta - 1 = \dfrac{1}{{{{\cos }^2}\phi }} - \dfrac{{{{\cos }^2}\phi }}{{{{\cos }^2}\phi }} \\
\Rightarrow 2{\cos ^2}\theta - 1 = \dfrac{1}{{{{\cos }^2}\phi }} - 1 \\
\Rightarrow 2{\cos ^2}\theta = \dfrac{1}{{{{\cos }^2}\phi }} \\
\Rightarrow {\cos ^2}\phi = \dfrac{1}{{2{{\cos }^2}\theta }} \\
\Rightarrow \cos \phi = \dfrac{1}{{\sqrt 2 \cos \theta }} \\
 $
Hence proved.

Note:
Here we know that $2\sin \theta \cos \theta = \sin 2\theta $ and also that ${\cos ^2}\theta - {\sin ^2}\theta = \cos 2\theta $ so we can use the formula ${\cos ^2}2\theta + {\sin ^2}2\theta = 1$ but we cannot use ${\cos ^2}\theta + {\sin ^2}2\theta = 1$ as $\theta {\text{ and }}\phi $ both are two different angles.