Question

tan ^-1 [3x^2 - 4y^2 / 3x^2 + 4y^2] = a^2​

Answer 1

Answer:

Step-by-step explanation:

$\tan ^-\left(1\right)\left[3x^2-4\cdot \frac{y^2}{3}x^2+4y^2\right]=a^2$

$\mathrm{Switch\:sides}$

$a^2=\tan ^-\left(1\right)\left[3x^2-4\cdot \frac{y^2}{3}x^2+4y^2\right]$

Multiply

$a^2=\tan ^-\left(1\right)\left(3x^2-\frac{4x^2y^2}{3}+4y^2\right)$

$\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$a=\sqrt{\tan ^-\left(1\right)\left(3x^2-\frac{4x^2y^2}{3}+4y^2\right)},\:a=-\sqrt{\tan ^-\left(1\right)\left(3x^2-\frac{4x^2y^2}{3}+4y^2\right)}$