Question

Find the general solution: \(\displaystyle{1}-\frac{{{\tan}^{{2}}{\left({x}\right)}}}{{1}}+{{\tan}^{{2}}{\left({x}\right)}}={1}-{2}{{\sin}^{{2}}{\left({x}\right)}}\)

Answer 1

\(\displaystyle{{\tan}^{{2}}{x}}=\frac{{{\sin}^{{2}}{x}}}{{{\cos}^{{2}}{x}}}{\quad\text{and}\quad}{{\cos}^{{2}}=}{1}-{{\sin}^{{2}}{x}}{\quad\text{and}\quad}{\cos}^{{2}}{x}+{{\sin}^{{2}}{x}}={1}\)
therefore \(\displaystyle{1}-\frac{{\frac{{{\sin}^{{2}}{x}}}{{{\cos}^{{2}}{x}}}}}{{1}}+{\left(\frac{{{\sin}^{{2}}{x}}}{{{\cos}^{{2}}{x}}}\right)}\)
then make the fractions of the top and bottom common by timeing by cosx therefore
\(\displaystyle\frac{{{{\cos}^{{2}}{x}}--{{\sin}^{{2}}{x}}}}{{{\cos}^{{2}}{x}+{{\sin}^{{2}}{x}}}}\)
therefore
\(\displaystyle{{\cos}^{{2}}{x}}-\frac{{{\sin}^{{2}}{x}}}{{1}}={{\cos}^{{2}}{x}}-{{\sin}^{{2}}{x}}\)
\(\displaystyle{{\cos}^{{2}}=}{1}-{{\sin}^{{2}}}\)
therefore
\(\displaystyle{\left({1}-{{\sin}^{{2}}{x}}\right)}-{{\sin}^{{2}}{x}}={1}-{2}{{\sin}^{{2}}{x}}\)