Question : If $0 \leq \theta \leq 90^{\circ}$ and $\sec ^{107} \theta+\cos ^{107} \theta=2$, then, $(\sec \theta+\cos \theta)$ is equal to:
Option 1: $2$
Option 2: $1$
Option 3: $\frac{1}{2}$
Option 4: $2^{-107}$
Correct Answer: $2$
Solution :
Given,
$\sec^{107} θ + \cos^{107} θ = 2$
Put $θ = 0°$
⇒ $\sec^{107} 0° + \cos^{107} 0° = 2$
⇒ $1^{107} + 1^{107} = 2$
⇒ 1 + 1 = 2
⇒ 2 = 2 (satisfied)
Now,
$\sec θ + \cos θ$
= $\sec 0° + \cos 0°$
= $1 + 1$
= $2$
Hence, the correct answer is $2$.
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