Question

If $\sec \theta -\tan \theta =5$ then find the value of $\theta$

Answer 1

$\sec \theta -\tan \theta =0-\left(i\right)$

we know that $\sec {}^2\theta -\tan {}^2\theta =1$

$\sec \theta +\tan \theta =\frac{1}{\sec \theta -\tan \theta }=\frac{1}{5}-\left(ii\right)$

Adding $\left(i\right)$ & $\left(ii\right)$ gives

$2\sec \theta =5+\frac{1}{5}2\sec \theta =\frac{26}{5}\sec \theta =\frac{13}{5}\cos \theta =\frac{5}{13}\theta ={\cos }^{-1}\left(\frac{5}{13}\right)$