Question

Tan@ = 2ab /a^2-b^2
Cos@=?

Answer 1

$\sin^2 \theta + \cos^2 \theta = 1$

$\implies \dfrac{\sin^2 \theta}{\cos^2 \theta} + 1 = \dfrac{1}{\cos^2 \theta}$

$\implies \tan^2 \theta + 1 = \dfrac{1}{\cos^2 \theta}$

 

$\tan \theta = \dfrac{2ab}{a^2-b^2}$

$\implies \dfrac{4a^2b^2}{a^4-2a^2b^2+b^4} + 1 = \dfrac{1}{\cos^2 \theta}$

$\implies \dfrac{a^4+2a^2b^2+b^4}{a^4-2a^2b^2+b^4} = \dfrac{1}{\cos^2 \theta}$

$\implies (\dfrac{a^2+b^2}{a^2-b^2})^2 = \dfrac{1}{\cos^2 \theta}$

$\therefore \cos \theta = \pm \dfrac{a^2-b^2}{a^2+b^2}$

 

The cosine has two values as it is unknown which quadrant $\theta$ lies.

However, the negative solution can be rejected if $\theta$ lies in the 1st Quadrant.