[ 1-tanA]^2/[1-cotA]^2=tan^A

1.	[ 1-tanA]^2/[1-cotA]^2=tan^A
Answer» {tex}\\begin{array}{l}\\frac{(1-\\mathrm{tanA})^2}{(1-\\mathrm{cotA})^2}=\\frac{(1-{\\displaystyle\\frac{\\mathrm{sinA}}{\\mathrm{cosA}}})^2}{(1-\\frac{\\mathrm{cosA}}{\\mathrm{sinA}})^2}\\\\=\\frac{{\\displaystyle(}\\cos{\\displaystyle\\mathrm A}{\\displaystyle-}{\\displaystyle\\mathrm{sinA}}{\\displaystyle{\\displaystyle)}^2}}{{\\displaystyle\\cos^2}{\\displaystyle\\mathrm A}}\\times\\frac{{\\displaystyle\\sin^2}{\\displaystyle\\mathrm A}}{(\\mathrm{sinA}-\\mathrm{cosA})^2}=\\frac{\\sin^2\\mathrm A}{{\\displaystyle\\cos}^2\\mathrm A}=\\tan^2\\mathrm A\\\\\\end{array}{/tex}

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