Question

If ${\cos }^2\theta -{\sin }^2\theta ={\tan }^2\varphi$ prove that :
$\cos \varphi =\frac{1}{\sqrt{2}\cos \theta }$

Answer 1

Given $({\cos }^2\theta -{\sin }^2\theta )={\tan }^2\varphi$

To prove:

$\cos \varphi =\frac{1}{\sqrt{2}\cos \theta }$

$({\cos }^2\theta -{\sin }^2\theta )={\tan }^2\varphi$

add $1$ on both sides

$({\cos }^2\theta -{\sin }^2\theta )={\tan }^2\varphi +1$

${\cos }^2+(1-\sin \theta )=(1+{\tan }^2\varphi )$

${\cos }^1\theta +{\cos }^2\theta ={\sec }^2\varphi$

$\Rightarrow 2{\cos }^2\theta ={\sec }^2\varphi$

$\Rightarrow 2{\cos }^2\theta {\cos }^2\varphi =1$

$\Rightarrow {\cos }^2\varphi =\frac{1}{2{\cos }^1\theta }$

$\Rightarrow \cos \theta =\frac{1}{\sqrt{2}\cos \theta }$