Saved Bookmarks
| 1. |
If sec A = x + 1/x then prove that secA .tan A = 2x or 1/2x |
| Answer» Given,\xa0sec\xa0{tex}\\theta{/tex}\xa0= x +\xa0{tex}\\frac{1}{4 x}{/tex}We know that, tan\xa0{tex}\\theta{/tex}\xa0=\xa0{tex}\\pm \\sqrt{\\sec ^{2} \\theta-1}{/tex}{tex}=\\pm \\sqrt{\\left(x+\\frac{1}{4 x}\\right)^{2}-1}{/tex}{tex}=\\pm \\sqrt{x^2 + \\frac{1}{(4x)^2}+2.x. \\frac{1}{(4x)}-1}{/tex}{tex}=\\pm \\sqrt{x^2 + \\frac{1}{(4x)^2}+ \\frac{1}{(2)}-1}{/tex}{tex}=\\pm \\sqrt{x^2 + \\frac{1}{(4x)^2}- \\frac{1}{(2)}}{/tex}{tex}=\\pm \\sqrt {\\left(x-\\frac{1}{4 x}\\right)^2}{/tex}{tex}=\\pm\\left(x-\\frac{1}{4 x}\\right){/tex}Now, sec\xa0{tex}\\theta{/tex}\xa0+ tan\xa0{tex}\\theta{/tex}\xa0=\xa0{tex}\\left(x+\\frac{1}{4 x}\\right) \\pm\\left(x-\\frac{1}{4 x}\\right){/tex}i)\xa0sec\xa0{tex}\\theta{/tex}\xa0+ tan\xa0{tex}\\theta{/tex}\xa0=\xa0{tex}\\left(x+\\frac{1}{4 x}\\right) +\\left(x-\\frac{1}{4 x}\\right){/tex}{tex}=2x{/tex}ii)\xa0sec\xa0{tex}\\theta{/tex}\xa0+ tan\xa0{tex}\\theta{/tex}\xa0=\xa0{tex}\\left(x+\\frac{1}{4 x}\\right) -\\left(x-\\frac{1}{4 x}\\right){/tex}{tex}=\\frac{1}{4x}-(-\\frac{1}{4x}) \\\\=\\frac{1}{4x}+\\frac{1}{4x}\\\\=\\frac{2}{4x}{/tex}{tex}=\\frac{1}{2 x}{/tex} | |