L.H.S\xa0= (1 + cotA + tanA) (sinA – cosA)= sinA – cosA + cotA sinA – cotA cosA + sinA tanA – tanA cosA{tex}= \\sin A – \\cos A + \\frac{{\\cos A}}{{\\sin A}} \\times \\sin A – \\cot A\\cos A + \\sin A\\;\\tan A – \\frac{{\\sin A}}{{\\cos A}} \\times \\cos A{/tex}= sinA – cosA + cosA – cotA cosA + sinA tanA – sinA= sinA tanA – cotA cosA……..(1)Now taking ;{tex}\\quad \\frac{{\\sec A}}{{\\cos e{c^2}A}} – \\frac{{\\cos ecA}}{{{{\\sec }^2}A}}{/tex}{tex} = \\frac{{\\frac{1}{{\\cos A}}}}{{\\frac{1}{{{{\\sin }^2}A}}}} – \\frac{{\\frac{1}{{\\sin A}}}}{{\\frac{1}{{{{\\cos }^2}A}}}}{/tex}{tex} = \\frac{{{{\\sin }^2}A}}{{\\cos A}} – \\frac{{{{\\cos }^2}A}}{{\\sin A}}{/tex}{tex} = \\sin A \\times \\frac{{\\sin A}}{{\\cos A}} – \\cos A \\times \\frac{{\\cos A}}{{\\sin A}}{/tex}{tex} = \\sin A \\times \\tan A – \\cos A \\times \\cot A{/tex}…….(2)From (1) & (2),(1 + cotA + tanA) (sinA – cosA) =\xa0{tex}\\frac { \\sec A } { cosec ^ { 2 } A } – \\frac { cosec A } { \\sec ^ { 2 } A }{/tex}\xa0= sinA.tanA – cosA.cotA\xa0Hence, Proved.

(1 + cotA + tanA) (sinA – cosA)= sinA – cosA + cotA sinA – cotA cosA + sinA tanA – tanA cosA{tex}= \\sin A – \\cos A + \\frac{{\\cos A}}{{\\sin A}} \\times \\sin A – \\cot A\\cos A + \\sin A\\;\\tan A – \\frac{{\\sin A}}{{\\cos A}} \\times \\cos A{/tex}= sinA – cosA + cosA – cotA cosA + sinA tanA – sinA= sinA tanA – cotA cosA……..(1)Now taking ;{tex}\\quad \\frac{{\\sec A}}{{\\cos e{c^2}A}} – \\frac{{\\cos ecA}}{{{{\\sec }^2}A}}{/tex}{tex} = \\frac{{\\frac{1}{{\\cos A}}}}{{\\frac{1}{{{{\\sin }^2}A}}}} – \\frac{{\\frac{1}{{\\sin A}}}}{{\\frac{1}{{{{\\cos }^2}A}}}}{/tex}{tex} = \\frac{{{{\\sin }^2}A}}{{\\cos A}} – \\frac{{{{\\cos }^2}A}}{{\\sin A}}{/tex}{tex} = \\sin A \\times \\frac{{\\sin A}}{{\\cos A}} – \\cos A \\times \\frac{{\\cos A}}{{\\sin A}}{/tex}{tex} = \\sin A \\times \\tan A – \\cos A \\times \\cot A{/tex}…….(2)From (1) & (2),(1 + cotA + tanA) (sinA – cosA) =\xa0{tex}\\frac { \\sec A } { cosec ^ { 2 } A } – \\frac { cosec A } { \\sec ^ { 2 } A }{/tex}\xa0= sinA.tanA – cosA.cotA\xa0