$ an\left\lbrack \frac{\pi}{2} + \frac{1}{2}\cos^{- 1}\left( \frac{a}{b} ight) ight brack + an\left\lbrack \frac{\pi}{4} - \frac{1}{2}\cos^{- 1}\left( \frac{a}{b} ight) ight brack$ is equal to

Question

$	an\left\lbrack \frac{\pi}{2} + \frac{1}{2}\cos^{- 1}\left( \frac{a}{b} ight) ightbrack + 	an\left\lbrack \frac{\pi}{4} - \frac{1}{2}\cos^{- 1}\left( \frac{a}{b} ight) ightbrack$ is equal to

Dr. Abubacker · Accepted Answer

(b)
$	herefore\ 	an\left\lbrack \frac{\pi}{4} + \frac{1}{2}\cos^{- 1}\left( \frac{a}{b} ight) ightbrack + 	an\left\lbrack \frac{\pi}{4} - \frac{1}{2}\cos^{- 1}\left( \frac{a}{b} ight) ightbrack$
$= 	an{\left\lbrack \frac{\pi}{4} + \phi ightbrack + 	an\left\lbrack \frac{\pi}{4} - \phi ightbrack}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left\lbrack put\frac{1}{2}\cos^{- 1}{\left( \frac{a}{b} ight) = \phi \Rightarrow \cos{2\phi = \frac{a}{b}}} ightbrack$
$= \frac{1 + 	an\phi}{1 - 	an\phi} + \frac{1 - 	an\phi}{1 + 	an\phi}$
$= \frac{2(1 + 	an^{2}\phi)}{1 - 	an^{2}\phi}$
$= \frac{2}{\cos{2\phi}} = \frac{2b}{a}$

Q. $\tan\left\lbrack \frac{\pi}{2} + \frac{1}{2}\cos^{- 1}\left( \frac{a}{b} \right) \right\rbrack + \tan\left\lbrack \frac{\pi}{4} - \frac{1}{2}\cos^{- 1}\left( \frac{a}{b} \right) \right\rbrack$ is equal to

Explanation

Reviewed by Dr. Abubacker

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