Thank you so much.
{tex}{{\\cos {\\rm{A}} – \\sin {\\rm{A}} + 1} \\over {\\cos {\\rm{A}} + \\sin {\\rm{A}} – 1}}{/tex}Dividing all terms by sin A, we get{tex}{{\\cot {\\rm{A}} – 1 + \\cos ec{\\rm{A}}} \\over {\\cot {\\rm{A}} + 1 – \\cos ec{\\rm{A}}}}{/tex}=\xa0{tex}{{\\cot {\\rm{A}} + \\cos ec{\\rm{A}} – 1} \\over {\\cot {\\rm{A}} – \\cos ec{\\rm{A}} + 1}}{/tex}=\xa0{tex}{{\\cot {\\rm{A}} + \\cos ec{\\rm{A}} – \\left( {\\cos e{c^2}{\\rm{A}} – {{\\cot }^2}{\\rm{A}}} \\right)} \\over {\\cot {\\rm{A}} – \\cos ec{\\rm{A}} + 1}}{/tex}=\xa0{tex}{{\\cot {\\rm{A}} + \\cos ec{\\rm{A}}\\left( {1 – \\cos ec{\\rm{A}} + \\cot {\\rm{A}}} \\right)} \\over {\\cot {\\rm{A}} – \\cos ec{\\rm{A}} + 1}}{/tex}=\xa0{tex}{\\cot {\\rm{A}} + \\cos ec{\\rm{A}}}{/tex}Hence proved
cosA-sinA+1/cosA+sinA-1= 1/sinA+cosA/sinAcosA-sinA+1/cosA+sinA-1=sinA+cosAsinA/sin (square)AcosA-sinA+1/cosA+sinA-1=sina(1+cosA)/sin(square)AcosA-sinA+1/cosA+sinA-1=1+cosA/sinAsinAcosA-sin(square)A+sinA=cosA+sinA-1+cos(square)A+sinAcosA-cosA(by cross multiplication)sincosA-sin(square)A+sinA=cosA+sinA-1+1-sin(square)A+sinAcosA-cosA0=0Hence proved
Yes

It\’s a NCERT book questions you can find the answer in mathematics solutions

By which identity
First divide each term by SinA and then solve it by trignometric identity
Plz solve this problem fastly

Thanks
Divide numerator and denominator by sinA OR By rationalize method u can solve this but first one is better than other

{tex}L H S=\\frac{\\cos A-\\sin A+1}{\\cos A+\\sin A-1}{/tex}{tex}=\\frac{\\sin A(\\cos A-\\sin A+1)}{\\sin A(\\cos A+\\sin A-1)}{/tex}{tex}=\\frac{\\sin A \\cos A-\\sin ^{2} A+\\sin A}{\\sin A(\\cos A+\\sin A-1)}{/tex}{tex}=\\frac{\\sin A \\cos A+\\sin A-\\left(1-\\cos ^{2} A\\right)}{\\sin A(\\cos A+\\sin A-1)}{/tex}{tex}=\\frac{\\sin A(\\cos A+1)-(1-\\cos A)(1+\\cos A)}{\\sin A(\\cos A+\\sin A-1)}{/tex}{tex}=\\frac{(1+\\cos A)(\\sin A+\\cos A-1)}{\\sin A(\\cos A+\\sin A-1)}{/tex}{tex}=\\frac{(1+\\cos A)(\\sin A+\\cos A-1)}{\\sin A(\\cos A+\\sin A-1)}{/tex}{tex}=\\frac{1}{\\sin A}+\\frac{\\cos A}{\\sin A}{/tex}= cos A + cot A = RHSProved

LHSDIVIDE NUMERATOR AND DENOMINATOR BOTH BY sin A =(CosA÷sinA – sinA÷sinA +1÷sinA)/(cosA÷sinA + sinA÷sinA – 1÷sinA)=(CotA-1+cosecA)/(cotA+1-cosecA)At the place of 1 at the numerator write (cosec2A -cot2A)={CotA+cosecA-(cosec2A-cot2A)}/(cotA-cosecA+1)={CotA+cosecA-(cosecA+cotA)(cosecA-cotA)}/(cotA-cosecA+1)=(CotA+cosecA){1-(cosecA-cotA)}/(cotA-cosecA+1)={(cotA+cosecA)(1-cosecA+cotA)}/(cotA-cosecA+1)=CotA+cosecAHENCE PROVEDLHS=RHS