Given: tan θ+ sec θ = l … eq. 1
Duplicating and isolating by (sec θ – tan θ) on numerator and denominator of L.H.S,
Along these lines, sec θ – tan θ = 1 … eq.2
Adding eq. 1and eq. 2, we get
![\[\left( tan\text{ }\theta \text{ }+\text{ }sec\text{ }\theta \right)\text{ }+\text{ }\left( sec\text{ }\theta \text{ }\text{ }tan\text{ }\theta \right)\text{ }=\text{ }1\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-8c449ab3cd97843b49475ee9ee47bb01_l3.png "Rendered by QuickLaTeX.com")
Henceforth, demonstrated.
If tanθ + secθ = l, then prove that secθ = (l2 + 1)/2l.