Question

$\frac{\alpha ^3}{2}cosec^2\left(\frac{1}{2}tan^{-1}\:\frac{\alpha }{\beta }\right)+\frac{\beta ^3}{2}sec^2\left(\frac{1}{2}tan^{-1}\:\frac{\beta }{\alpha }\right)\:=$

• 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,
$\frac{\alpha^{3}}{2}cosec^{2}\left(\frac{1}{2}tan^{-1} \frac{\alpha}{\beta}\right)+\frac{\beta^{3}}{2}sec^{2}\left(\frac{1}{2}tan^{-1} \frac{\beta}{\alpha}\right)$
$= \frac{\alpha ^{3}}{2\,sin^{2}\,\theta} + \frac{\beta ^{3}}{2\,cos^{2}\,\phi }$,
where $\theta = \frac{1}{2}tan^{-1} \frac{\alpha }{\beta }$
and $\phi = \frac{1}{2}tan^{-1} \frac{\beta }{\alpha }$
$= \frac{\alpha^{3}}{1 - cos\,2\theta} + \frac{\beta^{3}}{1 - cos\,2\phi}$
$= \frac{\alpha ^{3}}{1-cos\left(tan^{-1}\frac{\alpha }{\beta }\right)} + \frac{\beta ^{3}}{1+cos\left(tan^{-1}\frac{\beta }{\alpha }\right)}$
$= \frac{\alpha ^{3}}{1-cos\left(cos^{-1}\frac{\beta}{\sqrt{\alpha^{2}+\beta^{2}} }\right)}+\frac{\beta ^{3}}{1+cos\left(cos^{-1}\frac{\alpha }{\sqrt{\alpha ^{2}+\beta ^{2}} }\right)}$
$= \frac{\alpha ^{3}}{1-\frac{\beta }{\sqrt{\alpha ^{2}+\beta ^{2}} }}+\frac{\beta ^{3}}{1+\frac{\alpha }{\sqrt{\alpha ^{2}+\beta ^{2}} }}$
$= \left\{\frac{\alpha ^{3}}{\sqrt{\alpha ^{2}+\beta ^{2}} -\beta}+\frac{\beta ^{3}}{\sqrt{\alpha ^{2}+\beta ^{2}} +\alpha}\right\}\sqrt{\alpha ^{2}+\beta ^{2}}$
$= \left[\frac{\alpha^{3}\left\{\sqrt{\alpha ^{2}+\beta ^{2}}+\beta\right\}}{\alpha ^{2}+\beta ^{2} - \beta^{2}} +\frac{\beta ^{3}\left\{\sqrt{\alpha ^{2}+\beta ^{2}}-\alpha \right\}}{\alpha ^{2}+\beta ^{2} - \alpha ^{2}} \right]\sqrt{\alpha ^{2}+\beta ^{2}}$
$=\left\{\alpha\left(\sqrt{\alpha^{2}+\beta^{2}}+\beta\right)+\beta\left(\sqrt{\alpha^{2}+\beta^{2}}-\alpha\right)\right\}\sqrt{\alpha^{2}+\beta^{2}}$
$= \alpha\left(\alpha ^{2}+\beta ^{2}\right)+\beta\left(\alpha ^{2}+\beta ^{2}\right) = \left(\alpha+\beta\right)\left(\alpha ^{2}+\beta ^{2}\right)$