Question : If $\sin (a+b)=1$ and $\cos (a-b)=\frac{1}{2}$, find $a$.

Option 1: $75^{\circ}$

Option 2: $30^{\circ}$

Option 3: $15^{\circ}$

Option 4: $45^{\circ}$

Question : What will be the value of $\frac{\sin 30^{\circ} \sin 40^{\circ} \sin 50^{\circ} \sin 60^{\circ}}{\cos 30^{\circ} \cos 40^{\circ} \cos 50^{\circ} \cos 60^{\circ}}$?

Option 1: $\frac{1}{\sqrt{2}}$

Option 2: $\sqrt{3}$

Option 3: $1$

Option 4: $\frac{1}{\sqrt{3}}$

Question : Find the value of $\frac{\cos^2 15^{\circ}-\sin^2 15^{\circ}}{\cos^2 145^{\circ}+\sin^2 145^{\circ}}$.

Option 1: $\frac{1}{\sqrt{3}}$

Option 2: $\frac{1}{1-\sqrt{3}}$

Option 3: $\frac{\sqrt{3}}{2}$

Option 4: $\frac{2}{\sqrt{3}}$

Question : The value of $\frac{4 \tan ^2 30^{\circ}+\sin ^2 30^{\circ} \cos ^2 45^{\circ}+\sec ^2 48^{\circ}-\cot ^2 42^{\circ}}{\cos 37^{\circ} \sin 53^{\circ}+\sin 37^{\circ} \cos 53^{\circ}+\tan 18^{\circ} \tan 72^{\circ}}$ is:

Option 1: $\frac{35}{48}$

Option 2: $\frac{59}{48}$

Option 3: $\frac{49}{24}$

Option 4: $\frac{35}{24}$

Question : If $x\sin ^{2}60^{\circ}-\frac{3}{2}\sec 60^{\circ}\tan^{2}30^{\circ}+\frac{4}{5}\sin ^{2}45^{\circ}\tan ^{2}60^{\circ}=0$, then $x$ is:

Option 1: $-\frac{1}{15}$

Option 2: $–4$

Option 3: $-\frac{4}{15}$

Option 4: $–2$

Question : The value of $\left(\sin 30^{\circ} \cos 60^{\circ}-\cos 30^{\circ} \sin 60^{\circ}\right)$ is equal to:

Option 1: $-\cos 30^{\circ}$

Option 2: $-\sin 30^{\circ}$

Option 3: $\cos 30^{\circ}$

Option 4: $\sin 30^{\circ}$