Question

If $\tan \theta +\sec \theta =l$, then prove that $\sec \theta =\frac{l^2+l}{2l}$

Answer 1

Given,

$\tan \theta +\sec \theta =l$

$\tan \theta =l-\sec \theta$

squaring on both sides, we get,

${\tan }^2\theta =(l-\sec \theta {)}^2$

${\sec }^2\theta -1=l^2+{\sec }^2\theta -2l\sec \theta$

$l^2-2l\sec \theta =-1$

$l^2+1=2l\sec \theta$

$\Rightarrow \sec \theta =\frac{l^2+1}{2l}$