α32cosec2⁡12tan−1⁡αβ+β32sec2⁡12tan−1⁡βα is equal to

We have, α32cosec⁡12tan−1⁡αβ+β32sec⁡12tan−1⁡βα=α311−cos⁡tan−1⁡αβ+β11+cos⁡tan−1⁡βα =α11−cos⁡cos−1⁡βα2+β2+β311+cos⁡cos−1⁡1α2+β2 =α311−βα2+β2+β3⋅11+αα2+β2 =α2+β2α3α2+β2−β+β3α2+β2+β =α2+β2α3α2+β2+βα2+β3α2+β2−αβ2 =α2+β2αα2+β2+β+βα2+β2−α =α2+β2(α+β)α2+β2=(α+β)(α2+β2)

α32cosec2⁡12tan−1⁡αβ+β32sec2⁡12tan−1⁡βα is equal to

We have, α32cosec⁡12tan−1⁡αβ+β32sec⁡12tan−1⁡βα=α311−cos⁡tan−1⁡αβ+β11+cos⁡tan−1⁡βα =α11−cos⁡cos−1⁡βα2+β2+β311+cos⁡cos−1⁡1α2+β2 =α311−βα2+β2+β3⋅11+αα2+β2 =α2+β2α3α2+β2−β+β3α2+β2+β =α2+β2α3α2+β2+βα2+β3α2+β2−αβ2 =α2+β2αα2+β2+β+βα2+β2−α =α2+β2(α+β)α2+β2=(α+β)(α2+β2)

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$\frac{α^{3}}{2} {cosec}^{2} (\frac{1}{2} \tan^{- 1} \frac{α}{β}) + \frac{β^{3}}{2} \sec^{2} (\frac{1}{2} \tan^{- 1} \frac{β}{α})$ is equal to

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$(α - β) (α^{2} + β^{2})$

$(α + β) (α^{2} - β^{2})$

$(α + β) (α^{2} + β^{2})$

none of these

answer is C.

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Detailed Solution

We have, $\frac{α^{3}}{2} cosec (\frac{1}{2} \tan^{- 1} \frac{α}{β}) + \frac{β^{3}}{2} \sec (\frac{1}{2} \tan^{- 1} \frac{β}{α})$

$= α^{3} \frac{1}{1 - \cos (\tan^{- 1} \frac{α}{β})} + β \frac{1}{1 + \cos (\tan^{- 1} \frac{β}{α})} = α \frac{1}{1 - \cos (\cos^{- 1} \frac{β}{\sqrt{α^{2} + β^{2}}})} + β^{3} \frac{1}{1 + \cos (\cos^{- 1} \frac{1}{\sqrt{α^{2} + β^{2}}})} = α^{3} \frac{1}{1 - \frac{β}{\sqrt{α^{2} + β^{2}}}} + β^{3} \cdot \frac{1}{1 + \frac{α}{\sqrt{α^{2} + β^{2}}}} = \sqrt{α^{2} + β^{2}} [\frac{α^{3}}{\sqrt{α^{2} + β^{2}} - β} + \frac{β^{3}}{\sqrt{α^{2} + β^{2}} + β}] = \sqrt{α^{2} + β^{2}} [\frac{α^{3} (\sqrt{α^{2} + β^{2}} + β)}{α^{2}} + \frac{β^{3} (\sqrt{α^{2} + β^{2}} - α)}{β^{2}}] = \sqrt{α^{2} + β^{2}} [α (\sqrt{α^{2} + β^{2}} + β) + β (\sqrt{α^{2} + β^{2}} - α)] = \sqrt{α^{2} + β^{2}} [(α + β) (\sqrt{α^{2} + β^{2}})] = (α + β) (α^{2} + β^{2})$