-32cosec212tan-1-32212tan-1- is equal to

Α^32cosec^212(tan^ 1αβ)+β^32^2(12tan^ 1βα) =α^32sin^212(tan^ 1αβ)+β^32cos^212(tan^ 1βα) =α^31 cos(tan^ 1αβ)+β^31+cos(tan^ 1βα) Put tan^ 1αβ=cos^ 1β√(α^2+β^2) an ...

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Standard XII

Mathematics

Composition of Inverse Trigonometric Functions and Trigonometric Functions

α32cosec212ta...

Question

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

(α - β) (α^{2} + β^{2})

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(α + β) (α^{2} - β^{2})

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(α + β) (α^{2} + β^{2})

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None of the above

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Solution

The correct option is C $(α + β) (α^{2} + β^{2})$
$\frac{α^{3}}{2} {cosec}^{2} \frac{1}{2} ({tan}^{- 1} \frac{α}{β}) + \frac{β^{3}}{2} {sec}^{2} (\frac{1}{2} {tan}^{- 1} \frac{β}{α})$
$= \frac{α^{3}}{2 {sin}^{2} \frac{1}{2} ({tan}^{- 1} \frac{α}{β})} + \frac{β^{3}}{2 {cos}^{2} \frac{1}{2} ({tan}^{- 1} \frac{β}{α})}$
$= \frac{α^{3}}{1 - cos ({tan}^{- 1} \frac{α}{β})} + \frac{β^{3}}{1 + cos ({tan}^{- 1} \frac{β}{α})}$
Put ${tan}^{- 1} \frac{α}{β} = {cos}^{- 1} \frac{β}{\sqrt{α^{2} + β^{2}}}$ and ${tan}^{- 1} \frac{β}{α} = {cos}^{- 1} \frac{α}{\sqrt{α^{2} + β^{2}}}$
$= \frac{α^{3}}{1 - cos ({cos}^{- 1} \frac{β}{\sqrt{α^{2} + β^{2}}})} + \frac{β^{3}}{1 + cos ({cos}^{- 1} \frac{α}{\sqrt{α^{2} + β^{2}}})}$
$= \frac{α^{3}}{1 - \frac{β}{\sqrt{α^{2} + β^{2}}}} + \frac{β^{3}}{1 + \frac{α}{\sqrt{α^{2} + β^{2}}}}$
$= \sqrt{α^{2} + β^{2}} [\frac{α^{3}}{\sqrt{α^{2} + β^{2}} - β} + \frac{β^{3}}{\sqrt{α^{2} + β^{2}} + α}]$
$= \sqrt{α^{2} + β^{2}} [\frac{α^{3}}{\sqrt{α^{2} + β^{2}} - β} \times \frac{\sqrt{α^{2} + β^{2}} + β}{\sqrt{α^{2} + β^{2}} + β} + \frac{β^{3}}{\sqrt{α^{2} + β^{2}} + α} \times \frac{\sqrt{α^{2} + β^{2}} - α}{\sqrt{α^{2} + β^{2}} - α}]$
$= \sqrt{α^{2} + β^{2}} [α (\sqrt{α^{2} + β^{2}} + β) + β (\sqrt{α^{2} + β^{2}} - α)]$
$= \sqrt{α^{2} + β^{2}} [α \sqrt{α^{2} + β^{2}} + β \sqrt{α^{2} + β^{2}}]$
$= (α + β) (α^{2} + β^{2})$

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α32cosec212(tan−1αβ)+β32sec2(12tan−1βα) is equal to

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