Question : If $\sin(60^{\circ}-x)=\cos(y+60^{\circ})$, then the value of $\sin(x-y)$ is:
Option 1: $\frac{1}{\sqrt2}$
Option 2: $\frac{1}{2}$
Option 3: $\frac{\sqrt3}{2}$
Option 4: $1$
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Correct Answer: $\frac{1}{2}$
Solution : Given: $\sin(60^{\circ}-x)=\cos(y+60^{\circ})$ ⇒ $\cos(90^{\circ}-60^{\circ}+x)=\cos(y+60^{\circ})$ ⇒ $30^{\circ}+x=y+60^{\circ}$ ⇒ $x-y=30^{\circ}$ Now, $\sin(x-y)$ = $\sin(30^{\circ})$ = $\frac{1}{2}$ Hence, the correct answer is $\frac{1}{2}$.
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Question : If $\sin (x - y) = \frac{1}2$ and $\cos (x + y) = \frac{1}2$, then what is the value of $\sin x \cos x + 2\sin^2x + cos^3x \sec x$?
Option 1: $2$
Option 2: $\sqrt{2}+1$
Option 3: $1$
Option 4: $\frac{3}{4}$
Question : If $\tan (x + y)=1$ and $\cos (x-y)={\frac{\sqrt3}{2}}$, then what is the value of $x$ and $y$?
Option 1: $x = 3^\circ, y = 4.5^\circ$
Option 2: $x = 37.5^\circ, y = 7.5^\circ$
Option 3: $x = 7.5^\circ, y = 37.5^\circ$
Option 4: $x = 4.5^\circ, y = 3^\circ$
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 : If $\frac{x-x\tan^{2}30^{\circ}}{1+\tan^{2}30^{\circ}}=\sin^{2}30^{\circ}+4\cot^{2}45^{\circ}-\sec^{2}60^{\circ}$, then value of $x$ is:
Option 1: $\frac{1}{4}$
Option 2: $\frac{1}{5}$
Option 3: $\frac{1}{2}$
Option 4: $\frac{1}{\sqrt3}$
Question : If $\sec x+\cos x=\frac{5}{2}$, where $x$ lies between $0^{\circ}$ and $90^{\circ}$, then what is the value of $\sin ^2 x$ ?
Option 1: $\frac{3}{4}$
Option 4: $\frac{1}{4}$
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