InterviewSolution
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InterviewSolution
| 1. |
SecA(1_sinA)(secA+tanA)=1 |
| Answer» LHS\xa0{tex}= \\sec A(1 - \\sin A)(\\sec A + \\tan A){/tex}{tex} = (\\sec A - \\sec A \\times \\sin A)(\\sec A + \\tan A){/tex}{tex} = \\left( {\\sec A - \\frac{1}{{\\cos A}} \\times \\sin A} \\right)\\left( {\\sec A + \\tan A} \\right){/tex}\xa0{tex}\\left[ {\\because \\sec A = \\frac{1}{{\\cos A}}} \\right]{/tex}{tex} = \\left( {\\sec A - \\frac{{\\sin A}}{{\\cos A}}} \\right)(\\sec A + \\tan A){/tex}{tex} = (\\sec A - \\tan A)(\\sec A + \\tan A){/tex}\xa0{tex}\\left[ {\\because \\frac{{\\sin A}}{{\\cos A}} = \\tan A} \\right]{/tex}Using identity (a - b)(a + b) = a2 - b2= sec2 - tan2A= 1\xa0{tex}\\left[ {\\because {{\\sec }^2} - {{\\tan }^2}A = 1} \\right]{/tex}= RHSHence proved | |