JEE Main 2019TrigonometryTrigonometric Ratio and IdentitesEasyQuestionThe expression tanA1−cotA+cotA1−tanA{{\tan {\rm A}} \over {1 - \cot {\rm A}}} + {{\cot {\rm A}} \over {1 - \tan {\rm A}}}1−cotAtanA+1−tanAcotA can be written as:OptionsAsinA cosA+1\sin {\rm A}\,\cos {\rm A} + 1sinAcosA+1B secA cosecA+1\,\sec {\rm A}\,\cos ec{\rm A} + 1secAcosecA+1CtanA+cotA\tan {\rm A} + \cot {\rm A}tanA+cotADsecA+cosecA\sec {\rm A} + \cos ec{\rm A}secA+cosecACheck AnswerHide SolutionSolutionGiven expression can be written as sinAcosA×sin AsinA−cosA+cosAsinA×cosAcosA−sin A{{\sin A} \over {\cos A}} \times {{sin\,A} \over {\sin A - \cos A}} + {{\cos A} \over {\sin A}} \times {{\cos A} \over {\cos A - sin\,A}}cosAsinA×sinA−cosAsinA+sinAcosA×cosA−sinAcosA (As tanA=sinAcosA\tan A = {{\sin A} \over {\cos A}}tanA=cosAsinA and cotA=cosAsinA\cot A = {{\cos A} \over {\sin A}}cotA=sinAcosA ) =1sinA−cosA{sin3A−cos3AcosAsinA} = {1 \over {\sin A - \cos A}}\left\{ {{{{{\sin }^3}A - {{\cos }^3}A} \over {\cos A\sin A}}} \right\}=sinA−cosA1{cosAsinAsin3A−cos3A} =sin2A+sinAcosA+cos2 AsinAcosA = {{{{\sin }^2}A + \sin A\cos A + {{\cos }^2}\,A} \over {\sin A\cos A}}=sinAcosAsin2A+sinAcosA+cos2A =1+sec Acosec A = 1 + \sec\, A{\mathop{\rm cosec}\nolimits} \,A=1+secAcosecA