TanA+secA-1÷tanA-secA+1=1+sinA÷cosA
= tanA+secA-(sec^2A-tan^2)/tanA-secA+1 =tanA+secA-{(secA+tanA)(secA-tanA)/tanA-secA+1 =tanA+secA(1-secA+tanA)/tanA-secA+1 =tanA+secA =sinA/cosA + 1/cosA =1+sinA/cosA??i hope u understand
We have,{tex}\\mathrm { LHS } = \\frac { \\tan A } { 1 – \\cot A } + \\frac { \\cot A } { 1 – \\tan A }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan A } { 1 – \\frac { 1 } { \\tan A } } + \\frac { \\frac { 1 } { \\tan A } } { 1 – \\tan A }{/tex}{tex}\\Rightarrow \\quad \\text { LHS } = \\frac { \\tan A } { \\frac { \\tan A -1 } { \\tan A } } + \\frac { 1 } { \\tan A ( 1 – \\tan A ) }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A – 1 } + \\frac { 1 } { \\tan A ( 1 – \\tan A ) }{/tex}{tex}\\Rightarrow \\quad \\text { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A – 1 } – \\frac { 1 } { \\tan A ( \\tan A – 1 ) }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 3 } A – 1 } { \\tan A ( \\tan A – 1 ) }{/tex}\xa0[Taking LCM]{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { ( \\tan A – 1 ) \\left( \\tan ^ { 2 } A + \\tan A + 1 \\right) } { \\tan A ( \\tan A – 1 ) }{/tex}\xa0[{tex}\\because{/tex}\xa0a3\xa0- b3\xa0= ( a – b )(a2\xa0+ ab + b2)]{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A + \\tan A + 1 } { \\tan A }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A } + \\frac { \\tan A } { \\tan A } + \\frac { 1 } { \\tan A }{/tex}{tex}\\Rightarrow{/tex}\xa0LHS = tanA + 1 + cotA [ since (1/tanA) =cotA ].= (1 + tanA + cotA){tex}\\therefore \\quad \\frac { \\tan A } { 1 – \\cot A } + \\frac { \\cot A } { 1 – \\tan A }{/tex}\xa0= 1 + tanA + cotA ………..(1)Now, 1 + tanA + cotA = 1 +\xa0{tex}\\frac { \\sin A } { \\cos A } + \\frac { \\cos A } { \\sin A }{/tex}\xa0= 1 +\xa0{tex}\\frac { \\sin ^ { 2 } A + \\cos ^ { 2 } A } { \\sin A \\cos A }{/tex} = 1 +\xa0{tex}\\frac { 1 } { \\sin A \\cos A }{/tex}\xa0[{tex}\\because{/tex}Sin2A + Cos2 A = 1 ]\xa0= 1 + cosecAsecASo, 1 + tanA + cotA = 1+ cosecAsecA…….(2)From (1) and (2), we obtain{tex}\\frac { \\tan A } { 1 – \\cot A } + \\frac { \\cot A } { 1 – \\tan A }{/tex}\xa0= 1 + tanA + cotA = 1 + cosecAsecA