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| 1. |
SecA=2 X + 1 by 4 x prove that secant theta + tangent theta is equal to 2 x 2 x |
| Answer» By the given condition of question\xa0{tex}\\sec \\theta = x + \\frac { 1 } { 4 x }{/tex}{tex}\\therefore \\quad \\tan ^ { 2 } \\theta = \\sec ^ { 2 } \\theta - 1{/tex}{tex}\\Rightarrow \\quad \\tan ^ { 2 } \\theta = \\left( x + \\frac { 1 } { 4 x } \\right) ^ { 2 } - 1 = x ^ { 2 } + \\frac { 1 } { 16 x ^ { 2 } } + \\frac { 1 } { 2 } - 1 = x ^ { 2 } + \\frac { 1 } { 16 x ^ { 2 } } - \\frac { 1 } { 2 } = \\left( x - \\frac { 1 } { 4 x } \\right) ^ { 2 }{/tex}{tex}\\Rightarrow \\quad \\tan \\theta = \\pm \\left( x - \\frac { 1 } { 4 x } \\right){/tex}{tex}\\Rightarrow \\quad \\tan \\theta = \\left( x - \\frac { 1 } { 4 x } \\right) \\text { or, } \\tan \\theta = - \\left( x - \\frac { 1 } { 4 x } \\right){/tex}CASE 1: When\xa0{tex}\\tan \\theta = - \\left( x - \\frac { 1 } { 4 x } \\right) :{/tex}\xa0In this case,{tex}\\sec \\theta + \\tan \\theta = x + \\frac { 1 } { 4 x } + x - \\frac { 1 } { 4 x } = 2 x{/tex}CASE 2: When\xa0{tex}\\theta = - \\left( x - \\frac { 1 } { 4 x } \\right) :{/tex}\xa0In this case,{tex}\\sec \\theta + \\tan \\theta = \\left( x + \\frac { 1 } { 4 x } \\right) - \\left( x - \\frac { 1 } { 4 x } \\right) = \\frac { 2 } { 4 x } = \\frac { 1 } { 2 x }{/tex}Hence,\xa0{tex}\\sec \\theta + \\tan \\theta = 2 x \\text { or } , \\frac { 1 } { 2 x }{/tex} | |