General Aptitude TestQuantitative ReasoningTrigonometry

If 0 ≤ θ ≤ 90^°, and sec^107 θ + cos^107 θ = 2, then (sec θ + cos θ)…

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

Answer & explanation

Correct answer: option 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