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| 1. |
(1+cot°+tan°)(sin°-cos°)=sec°\\cosec°×cosec°-cosec°\\sec°×sec° |
| Answer» LHS = (1 + cot A + tan A) (sin A - cos A){tex}=\\left(1+\\frac{\\cos A}{\\sin A}+\\frac{\\sin A}{\\cos A}\\right){/tex}(sin A - cos A){tex}=\\frac{\\left(\\sin A \\cdot \\cos A+\\cos ^{2} A+\\sin ^{2} A\\right)}{\\sin A \\cdot \\cos A}{/tex}(sin A - cos A){tex}=\\frac{\\sin ^{3} A-\\cos ^{3} A}{\\sin A \\cdot \\cos A}{/tex}\xa0[Using (a\xa0-\xa0b) (a2\xa0+ b2\xa0+ ab) = [a3\xa0- b3]{tex}=\\frac{\\sin ^{3} A}{\\sin A \\cos A}-\\frac{\\cos ^{3} A}{\\sin A \\cdot \\cos A}{/tex}= sin2\xa0A\xa0{tex}\\times \\frac{1}{\\cos A}{/tex}\xa0- cos2\xa0A\xa0{tex}\\times \\frac{1}{\\sin A}{/tex}{tex}=\\frac{1}{\\ cosec ^{2} A} \\times{/tex}\xa0sec A\xa0{tex}-\\frac{1}{\\sec ^{2} A} \\times{/tex}\xa0cosec A{tex}=\\frac{\\sec A}{\\ cosec ^{2} A}-\\frac{\\ cosec A}{\\sec ^{2} A}{/tex}\xa0= RHS | |