Question : If $\cos(A+B)=0$ and $\sin(A-B)=0$, then what is the value of $\angle A$ and $\angle B$ ? ( $0^\circ < A, B \leq 90 ^\circ$)
Option 1: $20^\circ, 70^\circ$
Option 2: $45^\circ, 45^\circ$
Option 3: $60^\circ, 30^\circ$
Option 4: $15^\circ, 75^\circ$
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Correct Answer: $45^\circ, 45^\circ$
Solution : Given: $\cos(A+B)=0$ and $\sin(A-B)=0$ The equation $\cos(A+B)=\cos 90^\circ$ $A+B=90^\circ$............(equation 1) The equation $\sin(A-B)=\sin 0^\circ$ $A-B=0^\circ$............(equation 2) Solving these two equations, we get: $\angle A = \angle B = 45^\circ$ Hence, the correct answer is $45^\circ, 45^\circ$.
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Question : If $0^{\circ} < \theta < 90^{\circ}$ and $2 \sin^{2}\theta +3\cos\theta =3$, then the value of $\theta$ is:
Option 1: 30°
Option 2: 60°
Option 3: 45°
Option 4: 75°
Question : If $\sin (\theta +18^{\circ})=\cos 60^{\circ}(0< \theta < 90^{\circ})$, then the value of $\cos 5\theta$ is:
Option 1: $\frac{1}{2}$
Option 2: $0$
Option 3: $\frac{1}{\sqrt{2}}$
Option 4: $1$
Question : If $\cos A+\sin A=\sqrt{2}\cos A$, then $\cos A-\sin A$ is equal to: (where $0^{\circ}< A< 90^{\circ}$)
Option 1: $\sqrt{2}\sin A$
Option 2: $2\sin A$
Option 3: $2\sqrt{\sin A}$
Option 4: $\sqrt{2\sin A}$
Question : If $\frac{\cos \theta}{(1+\sin \theta)}+\frac{\cos \theta}{(1-\sin \theta)}=4$ and $\theta$ is acute, then the value of $\theta$ is:
Option 1: $60^{\circ}$
Option 2: $15^{\circ}$
Option 3: $45^{\circ}$
Option 4: $30^{\circ}$
Question : If $4\sin^{2}\theta-1=0$ and angle $\theta$ is less then $90^{\circ}$, the value of $\cos^{2}\theta+\tan^{2}\theta$ is: (Take $0^{\circ}< \theta< 90^{\circ}$)
Option 1: $\frac{17}{15}$
Option 2: $\frac{13}{12}$
Option 3: $\frac{11}{9}$
Option 4: $\frac{12}{11}$
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