Question

$\frac{{\alpha }^3}{2}cose{c}^2\left(\frac{1}{2}ta{n}^{-1}\frac{\alpha }{\beta }\right)+\frac{{\beta }^3}{2}se{c}^2\left(\frac{1}{2}ta{n}^{-1}\frac{\beta }{\alpha }\right)=$

• A. $\left(\alpha +\beta \right)\left({\alpha }^2+{\beta }^2\right)$
• B. $\left(\alpha -\beta \right)\left({\alpha }^2-{\beta }^2\right)$
• C. $\alpha +\beta$
• D. ${\alpha }^2+{\beta }^2$

Answer 1

Answer: (A)

We have,
$\frac{{\alpha }^3}{2}cose{c}^2\left(\frac{1}{2}ta{n}^{-1}\frac{\alpha }{\beta }\right)+\frac{{\beta }^3}{2}se{c}^2\left(\frac{1}{2}ta{n}^{-1}\frac{\beta }{\alpha }\right)$
$=\frac{{\alpha }^3}{2si{n}^2\theta }+\frac{{\beta }^3}{2co{s}^2\varphi }$,
where $\theta =\frac{1}{2}ta{n}^{-1}\frac{\alpha }{\beta }$
and $\varphi =\frac{1}{2}ta{n}^{-1}\frac{\beta }{\alpha }$
$=\frac{{\alpha }^3}{1-cos2\theta }+\frac{{\beta }^3}{1-cos2\varphi }$
$=\frac{{\alpha }^3}{1-cos\left(ta{n}^{-1}\frac{\alpha }{\beta }\right)}+\frac{{\beta }^3}{1+cos\left(ta{n}^{-1}\frac{\beta }{\alpha }\right)}$
$=\frac{{\alpha }^3}{1-cos\left(co{s}^{-1}\frac{\beta }{\sqrt{{\alpha }^2+{\beta }^2}}\right)}+\frac{{\beta }^3}{1+cos\left(co{s}^{-1}\frac{\alpha }{\sqrt{{\alpha }^2+{\beta }^2}}\right)}$
$=\frac{{\alpha }^3}{1-\frac{\beta }{\sqrt{{\alpha }^2+{\beta }^2}}}+\frac{{\beta }^3}{1+\frac{\alpha }{\sqrt{{\alpha }^2+{\beta }^2}}}$
$=\left\{\frac{{\alpha }^3}{\sqrt{{\alpha }^2+{\beta }^2}-\beta }+\frac{{\beta }^3}{\sqrt{{\alpha }^2+{\beta }^2}+\alpha }\right\}\sqrt{{\alpha }^2+{\beta }^2}$
$=\left[\frac{{\alpha }^3\left\{\sqrt{{\alpha }^2+{\beta }^2}+\beta \right\}}{{\alpha }^2+{\beta }^2-{\beta }^2}+\frac{{\beta }^3\left\{\sqrt{{\alpha }^2+{\beta }^2}-\alpha \right\}}{{\alpha }^2+{\beta }^2-{\alpha }^2}\right]\sqrt{{\alpha }^2+{\beta }^2}$
$=\left\{\alpha \left(\sqrt{{\alpha }^2+{\beta }^2}+\beta \right)+\beta \left(\sqrt{{\alpha }^2+{\beta }^2}-\alpha \right)\right\}\sqrt{{\alpha }^2+{\beta }^2}$
$=\alpha \left({\alpha }^2+{\beta }^2\right)+\beta \left({\alpha }^2+{\beta }^2\right)=\left(\alpha +\beta \right)\left({\alpha }^2+{\beta }^2\right)$