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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Question : If $\cot \theta=\frac{1}{\sqrt{3}}, 0^{\circ}<\theta<90^{\circ}$, then the value of $\frac{2-\sin ^2 \theta}{1-\cos ^2 \theta}+\left(\operatorname{cosec}^2 \theta-\sec \theta\right)$ is:
Question : If $0 \leq \theta \leq 90^{\circ}$, and $\sin \left(2 \theta+50^{\circ}\right)=\cos \left(4 \theta+16^{\circ}\right)$, then what is the value of $\theta$ (in degrees)?
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