JEE Main 2019TrigonometryTrigonometric Ratio and IdentitesEasyQuestionLet cos(α+β)=45\cos \left( {\alpha + \beta } \right) = {4 \over 5}cos(α+β)=54 and sin (α−β)=513,\sin \,\,\,\left( {\alpha - \beta } \right) = {5 \over {13}},sin(α−β)=135, where 0≤α, β≤π4.0 \le \alpha ,\,\beta \le {\pi \over 4}.0≤α,β≤4π. Then tan 2αtan\,2\alpha tan2α =OptionsA5633{56 \over 33}3356B1912{19 \over 12}1219C207{20 \over 7}720D2516{25 \over 16}1625Check AnswerHide SolutionSolutioncos(α+β)=45⇒tan(α+β)=34\cos \left( {\alpha + \beta } \right) = {4 \over 5} \Rightarrow \tan \left( {\alpha + \beta } \right) = {3 \over 4}cos(α+β)=54⇒tan(α+β)=43 sin(α−β)=513⇒tan(α−β)=512\sin \left( {\alpha - \beta } \right) = {5 \over {13}} \Rightarrow \tan \left( {\alpha - \beta } \right) = {5 \over {12}}sin(α−β)=135⇒tan(α−β)=125 tan2α=tan[(α+β)+(α−β)]\tan 2\alpha = \tan \left[ {\left( {\alpha + \beta } \right) + \left( {\alpha - \beta } \right)} \right]tan2α=tan[(α+β)+(α−β)] =34+5121−34.512=5633 = {{{3 \over 4} + {5 \over {12}}} \over {1 - {3 \over 4}.{5 \over {12}}}} = {{56} \over {33}}=1−43.12543+125=3356