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| 1. |
IF sec alpha =5/4 then evaluate tan alpha /(1+tan square alpha |
| Answer» We have,{tex} \\sec \\alpha = \\frac { \\text { Hypotenuse } } { \\text { Base } } = \\frac { 5 } { 4 }{/tex}So,Let us draw a right triangle ABC, {tex}\\angle B=90^\\circ{/tex}\xa0such thathypotenuse = AC = 5 units, Base = AB = 4 units, and{tex}\\angle B A C{/tex}= {tex}\\alpha{/tex}.Applying Pythagoras theorem in\xa0{tex}\\Delta{/tex}ABC, we get{tex}A C ^ { 2 } = A B ^ { 2 } + B C ^ { 2 }{/tex}{tex}\\Rightarrow \\quad 5 ^ { 2 } = 4 ^ { 2 } + B C ^ { 2 }{/tex}{tex}\\Rightarrow \\quad B C ^ { 2 } = 5 ^ { 2 } - 4 ^ { 2 } = 9{/tex}{tex}\\Rightarrow \\quad B C = \\sqrt { 9 } = 3{/tex}{tex}\\therefore \\quad \\tan \\alpha = \\frac { B C } { A B } = \\frac { 3 } { 4 }{/tex}Now, we have,\xa0{tex}\\frac { 1 - \\tan \\alpha } { 1 + \\tan \\alpha } = \\frac { 1 - \\frac { 3 } { 4 } } { 1 + \\frac { 3 } { 4 } } = \\frac { \\frac { 1 } { 4 } } { \\frac { 7 } { 4 } } = \\frac { 1 } { 7 }{/tex}therefore,\xa0{tex}\\frac{1-\\tan\\alpha}{1+\\tan\\alpha}=\\frac17{/tex} | |