RRB Group D Model Paper 3 with solutions

(3) We know that,
$\mathit{s}\mathit{e}{\mathit{c}}^2\mathit{\theta }-\mathit{t}\mathit{a}{\mathit{n}}^2\mathit{\theta }$ = 1
(secθ + tanθ ) (secθ - tanθ ) = 1
secθ - tanθ = $\frac{1}{\mathit{s}\mathit{e}\mathit{c}\mathit{\theta }+\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }}=\frac{1}{2}$
And, secθ + tanθ = 2 ;
therefore using the elimination method in both the equations.
2secθ = 2 + $\frac{1}{2}=\frac{5}{2}$ ⇒ secθ = $\frac{5}{4}$