Question

$\frac{{\alpha }^3}{2}cose{c}^2\frac{1}{2}\left({\tan }^{-1}\frac{\alpha }{\beta }\right)+\frac{{\beta }^2}{2}{\sec }^2\left(\frac{1}{2}{\tan }^{-1}\frac{\beta }{\alpha }\right)$ is equal to

• A. $(\alpha -\beta )({\alpha }^2+{\beta }^2)$
• B. $(\alpha +\beta )({\alpha }^2-{\alpha }^2)$
• C. $(\alpha +\beta )({\alpha }^2+{\beta }^2)$
• D. None of the above

Answer 1

The correct option is C $(\alpha +\beta )({\alpha }^2+{\beta }^2)$

We have,
$\frac{{\alpha }^3}{2{\sin }^2\left(\frac{1}{2}{\tan }^{-1}\frac{\alpha }{\beta }\right)}+\frac{{\beta }^3}{2{\cos }^2\left(\frac{1}{2}{\tan }^{-1}\frac{\beta }{\alpha }\right)}$
$=\frac{{\alpha }^3}{1-\cos \left({\tan }^{-1}\frac{\alpha }{\beta }\right)}+\frac{{\beta }^3}{1+\cos \left({\tan }^{-1}\frac{\beta }{\alpha }\right)}$
$=\frac{{\alpha }^3}{1-\cos \left({\cos }^{-1}\frac{\beta }{\sqrt{{\alpha }^2+{\beta }^2}}\right)}+\frac{{\beta }^3}{1+\cos \left({\cos }^{-1}\frac{\alpha }{\sqrt{{\alpha }^2+{\beta }^2}}\right)}$
$=\frac{{\alpha }^3}{1-\frac{\beta }{\sqrt{{\alpha }^2+{\beta }^2}}}+\frac{{\beta }^3}{1+\frac{\alpha }{\sqrt{{\alpha }^2+{\beta }^2}}}$
$=\sqrt{{\alpha }^2+{\beta }^2}[\frac{{\alpha }^3}{\sqrt{{\alpha }^2+{\beta }^2}-\beta }+\frac{{\beta }^3}{\sqrt{{\alpha }^2+{\beta }^2}+\alpha }]$
$=\sqrt{{\alpha }^2+{\beta }^2}[\frac{{\alpha }^3(\sqrt{{\alpha }^2+{\beta }^2}+\beta )}{{\alpha }^2+{\beta }^2-{\beta }^2}+\frac{{\beta }^3(\sqrt{{\alpha }^2+{\beta }^2}-\alpha )}{{\alpha }^2+{\beta }^2-{\alpha }^2}]$
$=\sqrt{{\alpha }^2+{\beta }^2}\left[\alpha \right(\sqrt{{\alpha }^2+{\beta }^2}+\beta )+\beta (\sqrt{{\alpha }^2+{\beta }^2}-\alpha \left)\right]$
$=\sqrt{{\alpha }^2+{\beta }^2}\left[\alpha \right(\sqrt{{\alpha }^2+{\beta }^2})+\beta (\sqrt{{\alpha }^2+{\beta }^2}\left)\right]$
$=({\alpha }^2+{\beta }^2)(\alpha +\beta )$.