Question

solve
$\tan \frac{1}{2} ( { \sin}^{ - 1} \frac{2x}{1 + {x}^{2} } + { \cos}^{ - 1} \frac{1 - {y}^{2} }{1 + {y}^{2} } )$
​

Answer 1

To solve :

$\star \: \tan \: \frac{1}{2} ( { \sin}^{ - 1} \frac{2x}{1 + {x}^{2} } + { \cos}^{ - 1} \frac{1 - {y}^{2} }{1 + {y}^{2} } )$

Solution :

•As we know that :

$\star \: \boxed{ \bold{ 2 \: { \tan}^{ - 1} x = { \sin}^{ - 1} \frac{2x}{1 + {x}^{2} } }} \\ \\ \star \boxed{ \bold{ 2 \: { \tan}^{ - 1} y = { \cos}^{ - 1} \frac{1 - {y}^{2} }{1 + {y}^{2} } }}$

$\implies \: \tan \: \frac{1}{2} (2 \: { \tan}^{ - 1} x + 2 { \tan}^{ - 1} y) \\ \\ \implies \: \tan \: \frac{1}{ \cancel {2}} \times \cancel{2}( { \tan}^{ - 1} x + { \tan}^{ - 1} y) \\ \\ \implies \: \tan \: ( { \tan}^{ - 1} x + { \tan}^{ - 1} y) \\ \\ \implies \: \tan( { \tan}^{ - 1} ( \frac{x + y}{1 - xy} )) \\ \\ \implies \: \frac{x +y }{1 - xy}$

Formula :

$\star \boxed{\bold{ { \tan}^{ - 1} x + { \tan}^{ - 1} y = { \tan}^{ - 1} ( \frac{x + y}{1 - xy} )}} \\ \\ \star \boxed{ \bold{ \tan ({ \tan}^{ - 1} \theta) = \theta}}$

Answer 2

$\huge\bf{\underbrace{\gray{TO\:FIND:-}}}$

• The value of $\bf{\tan{\dfrac{1}{2}}\:\Big[\:\sin^{-1}\:\dfrac{2x}{1\:+\:x^2}\:+\:\cos^{-1}\:\dfrac{1\:-\:y^2}{1\:+\:y^2}\:\Big]\:}$

$\huge\bf{\underbrace{\gray{SOLUTION:-}}}$

✞︎ We solve $\bf\red{\sin^{-1}\:\dfrac{2x}{1\:+\:x^2}\:}$ & $\bf\red{\cos^{-1}\:\dfrac{1\:-\:y^2}{1\:+\:y^2}\:}$ separately .

Solving $\bf{\underline{\gray{\sin^{-1}\:\dfrac{2x}{1\:+\:x^2}\:}}}$ :-

• $\bf\red{Putting}$, x = tanθ

$\rm{:\implies\:\sin^{-1}\:\Big(\dfrac{2\tan{\theta}}{1\:+\:\tan^2{\theta}}\Big)\:}$

• Using, 1 + tan²θ = sec²θ

$\rm{:\implies\:\sin^{-1}\:\Big(\dfrac{2\tan{\theta}}{\sec^2{\theta}}\Big)\:}$

$\rm{:\implies\:\sin^{-1}\:\Big(\dfrac{2\:\sin{\theta}/\cos{\theta}}{1/\cos^2{\theta}}\Big)\:}$

$\rm{:\implies\:\sin^{-1}\:\Big(2\:\dfrac{\sin{\theta}}{\cos{\theta}}\times{\cos^2{\theta}}\Big)\:}$

$\rm{:\implies\:\sin^{-1}\:(2\:\sin{\theta}.\cos{\theta})\:}$

• Using, 2 sinθ . cosθ = sin2θ

$\rm{:\implies\:\sin^{-1}\:(\sin2{\theta})\:}$

• Using, $\bf\red{\sin^{-1}\:(\sin2{\theta})\:}$ = 2θ

$\bf\green{:\implies\:2\theta}$

Solving $\bf{\underline{\gray{\cos^{-1}\:\dfrac{1\:-\:y^2}{1\:+\:y^2}\:}}}$ :-

• $\bf\red{Putting}$, y = $\bf\purple{\tan{\alpha}}$

$\rm{:\implies\:\cos^{-1}\:\Big(\dfrac{1\:-\:\tan^2{\alpha}}{1\:+\:\tan^2{\alpha}}\Big)\:}$

• Using, $\bf\purple{1\:+\:\tan^2{\alpha}\:=\:\sec^2{\alpha}\:}$

$\rm{:\implies\:\cos^{-1}\:\Big(\dfrac{1\:-\:\tan^2{\alpha}}{\sec^2{\alpha}}\Big)\:}$

$\rm{:\implies\:\cos^{-1}\:\Big(\dfrac{1\:-\:\sin^2{\alpha}/\cos^2{\alpha}}{1/\cos^2{\alpha}}\Big)\:}$

$\rm{:\implies\:\cos^{-1}\:\Big(\dfrac{\cos^2{\alpha}\:-\:\sin^2{\alpha}}{\cos^2{\alpha}}\:\times{\cos^2{\alpha}}\Big)\:}$

$\rm{:\implies\:\cos^{-1}\:(\cos^2{\alpha}\:-\:\sin^2{\alpha})\:}$

• Using, $\bf\purple{\cos^2{\alpha}\:-\:\sin^2{\alpha}\:=\:\cos2{\alpha}\:}$

$\rm{:\implies\:\cos^{-1}\:(\cos2{\alpha})\:}$

$\bf\green{:\implies\:2{\alpha}\:}$

☯︎ Now, we have

$\bf{\tan{\dfrac{1}{2}}\:\Big[\:\sin^{-1}\:\dfrac{2x}{1\:+\:x^2}\:+\:\cos^{-1}\:\dfrac{1\:-\:y^2}{1\:+\:y^2}\:\Big]\:}$

$\rm{:\implies\:\tan{\dfrac{1}{2}}\:[2\theta\:+\:2\alpha]\:}$

$\rm{:\implies\:\tan\:[\theta\:+\:\alpha]\:}$

$\rm{:\implies\:\dfrac{\tan{\theta}\:+\:\tan{\alpha}}{1\:-\:\tan{\theta}.\tan{\alpha}}\:}$

• Using,

1. x = tanθ
2. y = $\bf{\tan{\alpha}}$

$\bf\blue{:\implies\:\dfrac{x\:+\:y}{1\:-\:xy}\:}$