Question

If $\cos \theta =\frac{21}{29}$, then find the value of $\frac{\sec \theta }{\tan \theta -\sin \theta }$

Answer 1

$\cos \theta =\frac{21}{29}$

$\because {\sin }^2\theta +{\cos }^2\theta =1$

$\Rightarrow \sin \theta =\sqrt{-1{\left(\frac{21}{29}\right)}^2}=\frac{20}{29}$

now

$\frac{\sec \theta }{\tan \theta -\sin \theta }$

$=\frac{\frac{1}{\cos \theta }}{\frac{\sin \theta }{\cos \theta }-\sin \theta }$$=\frac{1}{\sin \theta (1-\cos \theta )}$

$\frac{1}{\frac{20}{29}(1-\frac{21}{29})}=\frac{1}{\frac{20}{29}\times \left(\frac{29-21}{29}\right)}$

$\frac{1}{\frac{20}{29}\times \frac{8}{29}}=\frac{29\times 29}{20\times 8}=\frac{841}{160}$