Question

Prove the following: 2tan⁻¹1/2+tan⁻¹1/7=tan⁻¹31/17

Answer 1

TO PROVE :–

$\\ { \bold{2 \: {tan}^{ - 1} ( \dfrac{1}{2} ) + {tan}^{ - 1}( \dfrac{1}{7} ) = {tan}^{ - 1}( \dfrac{31}{17}) }}$

SOLUTION :–

▪︎ Let's take L.H.S. –

$\\ { = \bold{2 \: {tan}^{ - 1} ( \dfrac{1}{2} ) + {tan}^{ - 1}( \dfrac{1}{7} ) }}$

• We know that –

$\\ \longrightarrow { \boxed{ \bold{2 \: {tan}^{ - 1}(x ) \: = {tan}^{ - 1} ( \frac{2x}{1 - {x}^{2} } )}}}$

• Applying formula –

$\\ { = \bold{ \: {tan}^{ - 1} [ \dfrac{2( \frac{1}{2} )}{1 - {( \frac{1}{2}) }^{2} } ] + {tan}^{ - 1}( \dfrac{1}{7} ) }}$

$\\ { = \bold{ \: {tan}^{ - 1} [ \dfrac{1}{1 - \frac{1}{4} } ] + {tan}^{ - 1}( \dfrac{1}{7} ) }}$

$\\ { = \bold{ \: {tan}^{ - 1} [ \dfrac{1}{ (\frac{3}{4} ) } ] + {tan}^{ - 1}( \dfrac{1}{7} ) }}$

$\\ { = \bold{ \: {tan}^{ - 1} ( \frac{4}{3} ) + {tan}^{ - 1}( \dfrac{1}{7} ) }}$

• We also know that –

$\\ \longrightarrow { \boxed{ \bold{ \: {tan}^{ - 1} (x ) + {tan}^{ - 1}( y ) = {tan}^{ - 1} [ \frac{x + y}{1 - xy} ] }}}$

• Applying formula –

$\\ \ { \bold{ = {tan}^{ - 1} [ \dfrac{ \frac{4}{3} + \frac{1}{7} }{1 - ( \frac{4}{3})( \frac{1}{7}) } ] }}$

$\\ \ { \bold{ = {tan}^{ - 1} [ \dfrac{ \frac{28 + 3}{21} }{1 - ( \frac{4}{21} ) } ] }}$

$\\ \ { \bold{ = {tan}^{ - 1} [ \dfrac{ \frac{31}{21} }{( \frac{21 - 4}{21} ) } ] }}$

$\\ \ { \bold{ = {tan}^{ - 1} [ \dfrac{ \frac{31}{ \cancel{21}} }{( \frac{17}{ \cancel{21}} ) } ] }}$

$\\ \ { \bold{ = {tan}^{ - 1} (\dfrac{31}{17} ) }}$

$\\ \ { \bold{ = R.H.S. \: \: \: \: \: (Hence \: \: proved) }}$