Question : In $\triangle{XYZ}$, right-angled at $Y$, if $\sin X = \frac{1}{2}$, find the value of $\cos X \cos Z + \sin X \sin Z$.
Option 1: $\frac{\sqrt{3}}{2}$
Option 2: $\frac{\sqrt{3}}{4}$
Option 3: $\frac{2}{\sqrt{3}}$
Option 4: $\sqrt{3}$
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Correct Answer: $\frac{\sqrt{3}}{2}$
Solution : Given: In $\triangle {XYZ}$, right-angled at $Y$. We know that the sum of all the angles in a triangle is 180°. $\sin X = \frac{1}{2}$ ⇒ $\sin X = \sin{30°}$ So, $\angle X = 30°$ Then $\angle Z = 60°$, because $\angle Y + \angle X + \angle Z = 180°$ Then, $\cos X \cos Z + \sin X \sin Z$ $=\cos 30° \cos 60° + \sin 30° \sin 60°$ $=\frac{\sqrt{3}}{2} × \frac{1}{2} + \frac{1}{2} × \frac{\sqrt{3}}{2}$ $ = \frac{\sqrt{3}}{2}$ Hence, the correct answer is $\frac{\sqrt{3}}{2}$.
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Question : In a right triangle for an acute angle $x$, if $\sin x=\frac{3}{7}$, then find the value of $\cos x$.
Option 1: $\frac{2}{7}$
Option 2: $\frac{3}{4}$
Option 3: $\frac{1}{\sqrt{3}}$
Option 4: $\frac{2\sqrt{10}}{7}$
Question : If $\cos x+\sin x=\sqrt{2} \cos x$, what is the value of $(\cos x-\sin x)^2+(\cos x+\sin x)^2$?
Option 1: $2$
Option 2: $1$
Option 3: $0$
Option 4: $\frac{1}{\sqrt{2}}$
Question : If $x\sin^{3}\theta +y\cos^{3}\theta=\sin\theta\cos\theta$ and $x\sin\theta-y\cos\theta=0$, then the value of $\left ( x^{2}+y^{2} \right )$ equals:
Option 1: $1$
Option 2: $\frac{1}{2}$
Option 3: $\frac{3}2$
Option 4: $2$
Question : Simplify the following: $\frac{\cos x-\sqrt{3} \sin x}{2}$
Option 1: $\cos \left(\frac{\pi}{3}-x\right)$
Option 2: $\sin \left(\frac{\pi}{3}+x\right)$
Option 3: $\cos \left(\frac{\pi}{3}+x\right)$
Option 4: $\sin \left(\frac{\pi}{3}-x\right)$
Question : If $\tan \frac{A}{2}=x$, then find $x$.
Option 1: $\frac{\sqrt{1+\cos A}}{\sqrt{1-\cos A}}$
Option 2: $\frac{\sqrt{1-\sin A}}{\sqrt{1+\cos A}}$
Option 3: $\frac{\sqrt{1-\cos A}}{\sqrt{1+\cos A}}$
Option 4: $\frac{\sqrt{\cos A-1}}{\sqrt{1+\cos A}}$
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