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| 1. |
(Cosec teta -cot teta)2 =1-cos teta ÷1+cos teta |
| Answer» LHS\xa0{tex}= \\frac{{1 - \\cos A}}{{1 + \\cos A}}{/tex}Multiplying numerator and denominator by 1 - cos A{tex}= \\frac{{(1 - \\cos A)(1 - \\cos A)}}{{(1 + \\cos A)(1 - \\cos A)}}{/tex}{tex}= \\frac{{{{(1 - \\cos A)}^2}}}{{1 - {{\\cos }^2}A}}{/tex}\xa0{tex} \\left[ {\\because (a + b)(a - b) = {a^2} - {b^2}} \\right]{/tex}{tex}= \\frac{{{{(1 - \\cos A)}^2}}}{{{{\\sin }^2}A}}{/tex}\xa0{tex} \\left[ {\\because 1 - {{\\cos }^2}A = {{\\sin }^2}A} \\right]{/tex}{tex}= {\\left( {\\frac{{1 - \\cos A}}{{\\sin A}}} \\right)^2}{/tex}{tex}= {\\left( {\\frac{1}{{\\sin A}} - \\frac{{\\cos A}}{{\\sin A}}} \\right)^2}{/tex}{tex}= {(\\cos ecA - \\cot A)^2}{/tex}\xa0{tex} \\left[ {\\because \\frac{1}{{\\sin A}} = \\cos ec\\;A,\\frac{{\\cos A}}{{\\sin A}} = \\cot A} \\right]{/tex}{tex}= {\\left[ { - 1(\\cot A - \\cos ecA)} \\right]^2}{/tex}= (cot A - cosec A)2Hence proved. | |