If $0 \leq \theta \leq 90^\circ$, and $\sec^{107} \theta + \cos^{107} \theta = 2$, then $(\sec \theta + \cos \theta)$ is equal to:

$2^{-107}$ 2 $\frac{1}{2}$ 1

2

We are given that :- sec107 θ + cos107 θ = 2 Let us assume that θ = 0º sec107 0º + cos107 0º = 2 1 + 1 = 2 2 = 2 LHS = RHS   ( satisfied ) So,  θ = 0º Now, ( secθ + cosθ ) = ( sec0º + cos0º )  = 1 + 1 = 2