Question

plz prove this question....​

Answer 1

Appropriate Question :-

$\rm :\longmapsto\:cos\bigg[ {tan}^{ - 1}\bigg(sin( {cot}^{ - 1}x)\bigg) \bigg] = \dfrac{ \sqrt{ {x}^{2} + 1}}{ \sqrt{ {x}^{2} + 2} }$

Solution :-

Consider,

$\rm :\longmapsto\:cos\bigg[ {tan}^{ - 1}\bigg(sin( {cot}^{ - 1}x)\bigg) \bigg]$

We know,

$\green{ \boxed{ \bf \: {cot}^{ - 1}x = {sin}^{ - 1}\dfrac{1}{ \sqrt{ {x}^{2} + 1} }}}$

[Proof attached in attachment]

$\rm \: = \: \:\:cos\bigg[ {tan}^{ - 1}\bigg(sin( {sin}^{ - 1}\dfrac{1}{ \sqrt{ {x}^{2} + 1 } } )\bigg) \bigg]$

We know,

$\green{ \boxed{ \bf \: {sin(sin}^{ - 1}x) = x}}$

$\rm \: = \: \:\:cos\bigg[ {tan}^{ - 1}\bigg(\dfrac{1}{ \sqrt{ {x}^{2} + 1 } } \bigg) \bigg]$

We know,

$\green{ \boxed{ \bf \: {tan}^{ - 1}x = {cos}^{ - 1}\dfrac{1}{ \sqrt{1 + {x}^{2} } } }}$

[Proof attached in attachment]

$\rm \: = \: \:\:cos\bigg[ {cos}^{ - 1}\bigg(\dfrac{ \sqrt{ {x}^{2} + 1} }{ \sqrt{ {x}^{2} + 2} } \bigg) \bigg]$

We know,

$\green{ \boxed{ \bf \: {cos(cos}^{ - 1}x) = x}}$

$\rm \: = \: \:\:\bigg[ \bigg(\dfrac{ \sqrt{ {x}^{2} + 1} }{ \sqrt{ {x}^{2} + 2} } \bigg) \bigg]$

Hence,

$\bf :\longmapsto\:cos\bigg[ {tan}^{ - 1}\bigg(sin( {cot}^{ - 1}x)\bigg) \bigg] = \dfrac{ \sqrt{ {x}^{2} + 1}}{ \sqrt{ {x}^{2} + 2} }$

Additional Information :-

$\green{ \boxed{ \bf \: {tan}^{ - 1}x + {tan}^{ - 1}y = {tan}^{ - 1}\bigg[\dfrac{x + y}{1 - xy} \bigg]}}$

$\red{ \boxed{ \bf \: {tan}^{ - 1}x - {tan}^{ - 1}y = {tan}^{ - 1}\bigg[\dfrac{x - y}{1 + xy} \bigg]}}$

$\green{ \boxed{ \bf \: {sin}^{ - 1}x + {sin}^{ - 1}y = {sin}^{ - 1}\bigg[x \sqrt{1 - {y}^{2}} + y \sqrt{1 - {x}^{2} } \bigg]}}$

$\red{ \boxed{ \bf \:{sin}^{ - 1}x - {sin}^{ - 1}y = {sin}^{ - 1}\bigg[x \sqrt{1 - {y}^{2}} - y \sqrt{1 - {x}^{2} } \bigg]}}$

$\green{ \boxed{ \bf \:{cos}^{ - 1}x - {cos}^{ - 1}y = {cos}^{ - 1}\bigg[xy + \sqrt{1 - {y}^{2}}\sqrt{1 - {x}^{2} } \bigg]}}$

$\red{ \boxed{ \bf \:{cos}^{ - 1}x + {cos}^{ - 1}y = {cos}^{ - 1}\bigg[xy - \sqrt{1 - {y}^{2}}\sqrt{1 - {x}^{2} } \bigg]}}$

Answer 2

here is it's diagrams

hence upper answer is correct

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