Verified answer

Question :-

[tex]\sf \: 2 {tan}^{ - 1} \dfrac{1}{2} - {tan}^{ - 1} x = \dfrac{\pi}{4} , \: then \: x = \\ \\ [/tex]

[tex]\large\underline{\sf{Solution-}}[/tex]

Given equation is

[tex]\sf \: 2 {tan}^{ - 1} \dfrac{1}{2} - {tan}^{ - 1} x = \dfrac{\pi}{4} \\ \\ [/tex]

can be rewritten as

[tex]\sf \: 2 {tan}^{ - 1} \dfrac{1}{2} - \dfrac{\pi}{4} = {tan}^{ - 1} x \\ \\ [/tex]

We know,

[tex]\boxed{ \sf{ \:2 {tan}^{ - 1} x = {tan}^{ - 1} \frac{2x}{1 - {x}^{2} } \: }} \\ \\ [/tex]

So, using this result, we get

[tex]\sf \: {tan}^{ - 1} \dfrac{2 \times \dfrac{1}{2} }{1 - {\bigg(\dfrac{1}{2} \bigg) }^{2} } - \frac{\pi}{4} = {tan}^{ - 1} x \\ \\ [/tex]

[tex]\sf \: {tan}^{ - 1} \dfrac{1}{1 - \dfrac{1}{4} } - \frac{\pi}{4} = {tan}^{ - 1} x \\ \\ [/tex]

[tex]\sf \: {tan}^{ - 1} \dfrac{1}{ \dfrac{4 - 1}{4} } - \frac{\pi}{4} = {tan}^{ - 1} x \\ \\ [/tex]

[tex]\sf \: {tan}^{ - 1} \dfrac{1}{ \dfrac{3}{4} } - \frac{\pi}{4} = {tan}^{ - 1} x \\ \\ [/tex]

[tex]\sf \: {tan}^{ - 1} \frac{4}{3} - \frac{\pi}{4} = {tan}^{ - 1} x \\ \\ [/tex]

We know,

[tex]\boxed{ \sf{ \:tan \frac{\pi}{4} = 1 \: \: \implies \: \frac{\pi}{4} = {tan}^{ - 1} 1 \: }} \\ \\ [/tex]

So, using this, above expression can be rewritten as

[tex]\sf \: {tan}^{ - 1} x = {tan}^{ - 1} \dfrac{4}{3} - {tan}^{ - 1} 1 \\ \\ [/tex]

We know,

[tex]\boxed{ \sf{ \: {tan}^{ - 1} x - {tan}^{ - 1} y = {tan}^{ - 1} \frac{x - y}{1 + xy} \: }} \\ \\ [/tex]

So, using this result, we get

[tex]\sf \: {tan}^{ - 1} x = {tan}^{ - 1} \dfrac{\dfrac{4}{3} - 1}{1 + 1 \times \dfrac{4}{3} } \\ \\ [/tex]

[tex]\sf \: {tan}^{ - 1} x = {tan}^{ - 1} \dfrac{\dfrac{4 - 3}{3}}{1 + \dfrac{4}{3} } \\ \\ [/tex]

[tex]\sf \: {tan}^{ - 1} x = {tan}^{ - 1} \dfrac{\dfrac{1}{3}}{ \dfrac{3 + 4}{3} } \\ \\ [/tex]

[tex]\sf \: {tan}^{ - 1} x = {tan}^{ - 1} \dfrac{1}{7} \\ \\ [/tex]

[tex]\bf\implies \:x = \dfrac{1}{7} \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information :-

[tex]\boxed{ \sf{ \:2 {tan}^{ - 1} x = {sin}^{ - 1} \frac{2x}{1 + {x}^{2} } \: }} \\ \\ [/tex]

[tex]\boxed{ \sf{ \:2 {tan}^{ - 1} x = {cos}^{ - 1} \frac{1 - {x}^{2} }{1 + {x}^{2} } \: }} \\ \\ [/tex]

[tex]\boxed{ \sf{ \:2{cos}^{ - 1}x = {cos}^{ - 1}( {2x}^{2} - 1) \: }} \\ \\ [/tex]

[tex]\boxed{ \sf{ \:2{cos}^{ - 1}x = {sin}^{ - 1}( 1 - {2x}^{2}) \: }} \\ \\ [/tex]

[tex]\boxed{ \sf{ \:2{sin}^{ - 1}x = {sin}^{ - 1}(2x \sqrt{1 - {x}^{2} } )\: }} \\ \\ [/tex]

[tex]\boxed{ \sf{ \:{sin}^{ - 1}x + {sin}^{ - 1}y = {sin}^{ - 1}(x \sqrt{1 - {y}^{2} } + y \sqrt{1 - {x}^{2} }) \: }} \\ \\ [/tex]

[tex]\boxed{ \sf{ \:{sin}^{ - 1}x - {sin}^{ - 1}y = {sin}^{ - 1}(x \sqrt{1 - {y}^{2} } - y \sqrt{1 - {x}^{2} }) \: }} \\ \\ [/tex]

Answer:

fine

Sister

sorry sorry sorry sorry sorry sorry sorry sorry sorry

nisha Sister