Question

If ${\displaystyle \tan \theta =\frac{1}{\sqrt{5}},\text{write the value of}\frac{(\cos e{c}^2\theta -{\sec }^2\theta )}{(\cos e{c}^2\theta -{\sec }^2\theta )}}$

Answer 1

   ${\displaystyle \frac{(\cos e{c}^2\theta -{\sec }^2\theta )}{(\cos e{c}^2\theta +{\sec }^2\theta )}}$

  =${\displaystyle \frac{(1+{\cot }^2\theta )-(1+{\tan }^2\theta )}{(1+{\cot }^2\theta )+(1+{\tan }^2\theta )}}$

    =${\displaystyle \frac{(1+\frac{1}{{\tan }^2\theta })-(1+{\tan }^2\theta )}{(1+\frac{1}{{\tan }^2\theta })-(1+{\tan }^2\theta )}}$

     =${\displaystyle \frac{(1+\frac{1}{{\tan }^2\theta }-1-{\tan }^2\theta )}{(1+\frac{1}{{\tan }^2\theta }+1+{\tan }^2\theta )}}$

      =${\displaystyle \frac{(\frac{1}{{\tan }^2\theta }-{\tan }^2\theta )}{(\frac{1}{{\tan }^2\theta }+{\tan }^2\theta +2)}}$

     =${\displaystyle \frac{{\left(\frac{\sqrt{5}}{1}\right)}^2-{\left(\frac{1}{\sqrt{5}}\right)}^2}{{\left(\frac{\sqrt{5}}{1}\right)}^2+{\left(\frac{1}{\sqrt{5}}\right)}^2+2}}$

    =${\displaystyle \frac{(\frac{5}{1}+\frac{1}{5})}{(\frac{5}{1}+\frac{1}{5}+\frac{2}{1})}}$

    =${\displaystyle \frac{\left(\frac{24}{5}\right)}{\left(\frac{36}{5}\right)}}$

    =${\displaystyle \frac{24}{36}}$

     =${\displaystyle \frac{2}{3}}$