2{tan ^{ - 1}}left[ {sqrt {frac{{a - b}}{{a + | vedclass

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(A) We use the formula $2{	an ^{ - 1}}x = {\cos ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} ight)$.Let $x = \sqrt {\frac{{a - b}}{{a + b}}}

Vedclass · Accepted Answer

${\cos ^{ - 1}}\left( {\frac{{a\cos 	heta + b}}{{a + b\cos 	heta }}} ight)$

Vedclass · Answer

(A) We use the formula $2{	an ^{ - 1}}x = {\cos ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} ight)$.Let $x = \sqrt {\frac{{a - b}}{{a + b}}} 	an \frac{	heta }{2}$.Then $2{	an ^{ - 1}}\left[ {\sqrt {\frac{{a - b}}{{a + b}}} 	an \frac{	heta }{2}} ight] = {\cos ^{ - 1}}\left[ {\frac{{1 - \left( {\frac{{a - b}}{{a + b}}} ight){{	an }^2}\frac{	heta }{2}}}{{1 + \left( {\frac{{a - b}}{{a + b}}} ight){{	an }^2}\frac{	heta }{2}}}} ight]$$= {\cos ^{ - 1}}\left[ {\frac{{(a + b) - (a - b){{	an }^2}\frac{	heta }{2}}}{{(a + b) + (a - b){{	an }^2}\frac{	heta }{2}}}} ight]$$= {\cos ^{ - 1}}\left[ {\frac{{a(1 - {{	an }^2}\frac{	heta }{2}) + b(1 + {{	an }^2}\frac{	heta }{2})}}{{a(1 + {{	an }^2}\frac{	heta }{2}) + b(1 - {{	an }^2}\frac{	heta }{2})}}} ight]$Dividing numerator and denominator by $(1 + {	an ^2}\frac{	heta }{2})$:$= {\cos ^{ - 1}}\left[ {\frac{{a\left( {\frac{{1 - {{	an }^2}\frac{	heta }{2}}}{{1 + {{	an }^2}\frac{	heta }{2}}}} ight) + b}}{{a + b\left( {\frac{{1 - {{	an }^2}\frac{	heta }{2}}}{{1 + {{	an }^2}\frac{	heta }{2}}}} ight)}}} ight]$Using the identity $\cos 	heta = \frac{{1 - {{	an }^2}\frac{	heta }{2}}}{{1 + {{	an }^2}\frac{	heta }{2}}}$,we get:$= {\cos ^{ - 1}}\left( {\frac{{a\cos 	heta + b}}{{a + b\cos 	heta }}} ight)$.

$2{\tan ^{ - 1}}\left[ {\sqrt {\frac{{a - b}}{{a + b}}} \tan \frac{\theta }{2}} \right] = $

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