Question

Hey Mate !


If ,

( a² - b² ) sin α + 2ab cos α = a² + b².

Prove that :

tan α = ( a² - b² ) / 2ab

Try to use the easiest way .

Answer 1

Given Equation is (a^2 - b^2) sina + 2abcosa = a^2 + b^2.

It can be written as:

$= \ \textgreater \ \frac{a^2 - b^2}{a^2 + b^2} sina + \frac{2ab}{a^2 + b^2} cosa = 1$

On Equating the corresponding terms, we get

= > sina = (a^2 - b^2)/(a^2 + b^2)

= > cosa = (2ab)/(a^2 + b^2)


Therefore:

= > Tana = sina/cosa

               $= \frac{a^2 - b^2}{a^2 + b^2} * \frac{a^2 + b^2}{2ab}$

               $= \frac{a^2 - b^2}{2ab}$



Hope this helps!