Question

Which of the following is equal to \({\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{8{\rm{x}}\sqrt {\rm{x}} }}{{1 - 16{{\rm{x}}^3}}}} \right)\). 
1. \(2{\rm{ta}}{{\rm{n}}^{ - 1}}\left( {4{\rm{x}}\sqrt {\rm{x}} } \right)\)
2. \({\rm{ta}}{{\rm{n}}^{ - 1}}\left( {8{\rm{x}}\sqrt {\rm{x}} } \right)\)
3. \({\rm{ta}}{{\rm{n}}^{ - 1}}\left( {4{\rm{x}}\sqrt {\rm{x}} } \right)\)
4. \({\rm{ta}}{{\rm{n}}^{ - 1}}\left( {4{\rm{x}}} \right)\)
5. None of these

Answer 1

Correct Answer - Option 1 : \(2{\rm{ta}}{{\rm{n}}^{ - 1}}\left( {4{\rm{x}}\sqrt {\rm{x}} } \right)\)

Concept:

1. Inverse trigonometric functions:

• \({\rm{ta}}{{\rm{n}}^{ - 1}}{\rm{x}} - {\rm{ta}}{{\rm{n}}^{ - 1}}{\rm{y}} = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{{\rm{x}} - {\rm{y}}}}{{1 + {\rm{xy}}}}} \right)\)
• \({\rm{ta}}{{\rm{n}}^{ - 1}}{\rm{x}} + {\rm{ta}}{{\rm{n}}^{ - 1}}{\rm{y}} = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{{\rm{x}} + {\rm{y}}}}{{1 - {\rm{xy}}}}} \right)\)
• \(2{\rm{ta}}{{\rm{n}}^{ - 1}}{\rm{x}} = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{2{\rm{x}}}}{{1 - {{\rm{x}}^2}}}} \right)\)

Calculation:

Rewrite the given equation as follows:

\({\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{8{\rm{x}}\sqrt {\rm{x}} }}{{1 - 16{{\rm{x}}^3}}}} \right) = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{2\left( {4{\rm{x}}\sqrt {\rm{x}} } \right)}}{{1 - {\rm{\;}}{{\left( {4{\rm{x}}\sqrt {\rm{x}} } \right)}^2}}}} \right)\)

\(= 2{\rm{ta}}{{\rm{n}}^{ - 1}}\left( {4{\rm{x}}\sqrt {\rm{x}} } \right)\)