Question

Simplify tan 1/2[sin^-1 2x/1+x² + cos ^-1 1-y²/1+y²]​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:tan \: \dfrac{1}{2} \bigg( {sin}^{ - 1} \dfrac{2x}{1 + {x}^{2} } + {cos}^{ - 1}\dfrac{1 - {y}^{2} }{1 + {y}^{2} } \bigg)$

To simply this, we substitute x = tanp and y = tanq

$\rm \: = \: \: \:tan \: \dfrac{1}{2}\bigg( {sin}^{ - 1} \dfrac{2tanp}{1 + {tan}^{2} p} + {cos}^{ - 1}\dfrac{1 - {tan}^{2}q }{1 + {tan}^{2} q} \bigg)$

We know,

$\underbrace{\boxed{ \tt{sin2x = \frac{2tanx}{1 + {tan}^{2}x }\: }}}$

and

$\underbrace{\boxed{ \tt{cos2x = \frac{1 - {tan}^{2} x}{1 + {tan}^{2}x }\: }}}$

So, using this we get

$\rm \: = \: \: \:tan \: \dfrac{1}{2}\bigg( {sin}^{ - 1} (sin2p) + {cos}^{ - 1}(cos2q)\bigg)$

We know,

$\underbrace{\boxed{ \tt{ {sin}^{ - 1}(sinx) \: = \: x }}}$

and

$\underbrace{\boxed{ \tt{ {cos}^{ - 1}(cosx) \: = \: x }}}$

So, using this we get

$\rm \: = \: \: tan\dfrac{1}{2}(2p + 2q)$

$\rm \: = \: \: tan(p + q)$

$\rm \: = \: \: \dfrac{tanp + tanq}{1 - tanp \: tanq}$

$\rm \: = \: \: \dfrac{x + y}{1 - xy}$

Hence,

$\boxed{\rm\:tan \: \dfrac{1}{2} \bigg( {sin}^{ - 1} \dfrac{2x}{1 + {x}^{2} } + {cos}^{ - 1}\dfrac{1 - {y}^{2} }{1 + {y}^{2} } \bigg) = \frac{x + y}{1 - xy}}$

Additional Information :-

$\boxed{ \sf{ \: {tan}^{ - 1}x + {tan}^{ - 1}y = {tan}^{ - 1} \frac{x + y}{1 - xy}}}$

$\boxed{ \sf{ \: {tan}^{ - 1}x - {tan}^{ - 1}y = {tan}^{ - 1} \frac{x - y}{1 + xy}}}$

$\boxed{ \sf{ \: {sin}^{ - 1}x + {sin}^{ - 1}y = {sin}^{ - 1}(x \sqrt{1 - {y}^{2} } + y \sqrt{1 - {x}^{2} }}}$

$\boxed{ \sf{ \: {sin}^{ - 1}x - {sin}^{ - 1}y = {sin}^{ - 1}(x \sqrt{1 - {y}^{2} } - y \sqrt{1 - {x}^{2} }}}$

$\boxed{ \sf{ \: {cos}^{ - 1}x - {cos}^{ - 1}y = {cos}^{ - 1}(xy + \sqrt{1 - {y}^{2} } \sqrt{1 - {x}^{2} }}}$

$\boxed{ \sf{ \: {cos}^{ - 1}x + {cos}^{ - 1}y = {cos}^{ - 1}(xy - \sqrt{1 - {y}^{2} } \sqrt{1 - {x}^{2} }}}$