$ an\left\lbrack \frac{1}{2}\sin^{- 1}\left( \frac{2a}{1 + a^{2}} ight) + \frac{1}{2}\cos^{- 1}\left( \frac{1 - a^{2}}{1 + a^{2}} ight) ight brack\$is equal to

Question

$	an\left\lbrack \frac{1}{2}\sin^{- 1}\left( \frac{2a}{1 + a^{2}} ight) + \frac{1}{2}\cos^{- 1}\left( \frac{1 - a^{2}}{1 + a^{2}} ight) ightbrack\$is equal to

Dr. Abubacker · Accepted Answer

(c)
$	an\left\lbrack \frac{1}{2}\sin^{- 1}\ \left( \frac{2a}{1 + a^{2}} ight) + \frac{1}{2}cos^{- 1}\left( \frac{1 - a^{2}}{1 + a^{2}} ight) ightbrack$
$= 	an\left\lbrack \frac{1}{2} \bullet 2	an^{- 1}{a +}\frac{1}{2} \bullet 2	an^{- 1}a ightbrack$
$= tan(2	an^{- 1}{a)}$
$= 	an\left\lbrack 	an^{- 1}\left( \frac{2a}{1 - a^{2}} ight) ightbrack$
$= \frac{2a}{1 - a^{2}}$

Q. $\tan\left\lbrack \frac{1}{2}\sin^{- 1}\left( \frac{2a}{1 + a^{2}} \right) + \frac{1}{2}\cos^{- 1}\left( \frac{1 - a^{2}}{1 + a^{2}} \right) \right\rbrack\$is equal to

Explanation

Reviewed by Dr. Abubacker

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