Question

if ${\displaystyle \tan \theta =\frac{1}{\sqrt{2}}}$ find the value of ${\displaystyle \frac{\cos e{c}^2\theta -{\sec }^2\theta }{\cos e{c}^2\theta +{\cot }^2\theta }}$

Answer 1

Given ${\displaystyle \tan \theta =\frac{1}{\sqrt{2}}}$

We have to find the value of the expression ${\displaystyle \frac{\cos e{c}^2\theta -{\sec }^2\theta }{\cos e{c}^2\theta +{\cot }^2\theta }}$

We know that,

${\displaystyle 1+{\cot }^2\theta =\cos e{c}^2\theta }$

${\displaystyle \Rightarrow \cos e{c}^2\theta -{\cot }^2\theta =1}$

Therefore, the given expression can be written as

${\displaystyle \frac{\cos e{c}^2\theta -{\sec }^2\theta }{\cos e{c}^2\theta +{\cot }^2\theta }=\frac{\cos e{c}^2\theta -{\sec }^2\theta }{1+{\cot }^2\theta +{\cot }^2\theta }}$

${\displaystyle \tan \theta =\frac{1}{\sqrt{2}}\Rightarrow \cot \theta =\sqrt{2}}$

${\displaystyle \frac{\cos e{c}^2\theta -{\sec }^2\theta }{1+2{\cot }^2\theta }=\frac{1+{\cot }^2\theta - (1+{\tan }^2\theta )}{1+2\cot 62\theta }}$             (since ${\displaystyle {\sec }^2\theta =1+{\tan }^2\theta }$)

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

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

${\displaystyle =\frac{3}{10}}$

Hence, the value of the given expression is 3/10