Question : The value of $\cos \ 0°+\cos\ 1°+\cos\ 2°......\cos\ 180°$ is:
Option 1: $0$
Option 2: $1$
Option 3: $\frac{\sqrt3}{2}$
Option 4: $\frac{1}{2}$
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Correct Answer: $0$
Solution : Given: $\cos 0°+\cos 1°+\cos 2°......\cos 180°$ We know the trigonometric identity, $\cos \theta+\cos (180°–\theta)=0$. $\cos 0°+\cos 1°+\cos 2°......\cos 180°= (\cos 0°+\cos 180°)+(\cos\ 1°+\cos 179°)......(\cos 89°+\cos 91°)+\cos\ 90°= 0$ Hence, the correct answer is $0$.
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Question : If $\cos 20°= m$ and $\cos 70°=n$, then the value of $m^2+n^2$:
Option 1: $1$
Option 2: $\frac{3}{2}$
Option 3: $\frac{1}{\sqrt2}$
Question : What is the value of $\frac{3 \operatorname{cosec} 42°}{\sec 48°}-\frac{5 \cos 32°}{\sin 58°}$?
Option 1: - 1
Option 2: 5
Option 3: 0
Option 4: - 2
Question : $\triangle$XYZ is right angled at Y. If $\angle$X = 60°, then find the value of $(\sec Z+\frac{2}{\sqrt3})$.
Option 1: $\frac{4}{\sqrt3}$
Option 2: $\frac{\sqrt2+2}{2\sqrt2}$
Option 3: $\frac{7}{2\sqrt3}$
Option 4: $\frac{4}{2\sqrt3}$
Question : What is the value of $\sec 330°$?
Option 1: $2$
Option 2: $\frac{–2}{\sqrt3}$
Option 3: $–2$
Option 4: $\frac{2}{\sqrt3}$
Question : The value of $\text{cosec}^{2}\: 18°-\frac{1}{\cot^{2}72°}$ is:
Option 1: $\frac{1}{\sqrt3}$
Option 2: $\frac{\sqrt2}{3}$
Option 3: $\frac{1}{2}$
Option 4: $1$
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