Question

$\frac{a^{3}}{2}co\text{se}\text{c}^{2}\left( \frac{1}{2}\tan^{- 1}\frac{a}{b} \right) + \frac{b^{3}}{2}\sec^{2}\left( \frac{1}{2}\tan^{- 1}\frac{b}{a} \right)$ is equal to

• A. $(a - b)(a^{2} + b^{2})$
• B. $(a + b)(a^{2} - b^{2})$
• C. $(a + b)(a^{2} + b^{2})$
• D. None of these

Answer 1

Answer: (C)

$(a + b)(a^{2} + b^{2})$

Answer 2

Let $\tan^{- 1}\frac{a}{b} = \theta,\tan^{- 1}\frac{b}{a} = \varphi$ ⇒ $\therefore$ $\tan\theta = \frac{a}{b},\tan\varphi = \frac{b}{a}$

$\frac{a^{3}}{2}c\text{ose}\text{c}^{2}\left( \frac{1}{2}\tan^{- 1}\frac{a}{b} \right) + \frac{b^{3}}{2}\sec^{2}\left( \frac{1}{2}\tan^{- 1}\frac{b}{a} \right)$

=$\frac{a^{3}}{2\sin^{2}\left( \frac{\theta}{2} \right)}$+$\frac{b^{3}}{2\cos^{2}\left( \frac{\varphi}{2} \right)}$

=$\frac{a^{3}}{1 - \cos\theta} + \frac{b^{3}}{1 + \cos\varphi} = \frac{a^{3}}{1 - \frac{b}{\sqrt{a^{2} + b^{2}}}} + \frac{b^{3}}{1 + \frac{a}{\sqrt{a^{2} + b^{2}}}}$= $\sqrt{a^{2} + b^{2}}\left\lbrack \frac{a^{3}\lbrack\sqrt{a^{2} + b^{2}} + b\rbrack}{(a^{2} + b^{2}) - b^{2}} + \frac{b^{3}\lbrack\sqrt{a^{2} + b^{2}} - a\rbrack}{(a^{2} + b^{2}) - a^{2}} \right\rbrack$ (rationalized)

$= \sqrt{a^{2} + b^{2}}\lbrack a\{\sqrt{a^{2} + b^{2}} + b\} + b\{\sqrt{a^{2} + b^{2}} - a\}\rbrack$

= $\sqrt{a^{2} + b^{2}}\lbrack\sqrt{a^{2} + b^{2}}(a + b)\rbrack = (a^{2} + b^{2})(a + b)$