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: $\sqrt{2\sin A}$
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Correct Answer: $\sqrt{2}\sin A$
Solution : $\cos A+\sin A = \sqrt{2}\cos A$ $⇒\sin A = (\sqrt{2}-1)\cos A$ $⇒\frac {\sin A}{\cos A}=\sqrt{2}-1$ $⇒\frac {\sin A}{\cos A}=(\sqrt{2}-1)×\frac{(\sqrt{2}+1)}{(\sqrt{2}+1)}$ $⇒\frac {\sin A}{\cos A}=\frac{1}{\sqrt{2}+1}$ $⇒\cos A=(\sqrt{2}+1)\sin A$ $⇒\cos A=\sqrt{2}\sin A+\sin A$ $⇒\cos A - \sin A = \sqrt{2}\sin A$ Hence, the correct answer is $\sqrt{2}\sin A$.
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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 $\sin 2\theta=\frac{\sqrt{3}}{2}$, then the value of $\sin 3\theta$ is equal to $(0^{\circ}\leq \theta\leq 90^{\circ})$:
Option 2: $1$
Option 3: $0$
Option 4: $\frac{\sqrt{3}}{2}$
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 $\theta$ is a positive acute angle and $4\cos ^{2}\theta -4\cos \theta +1=0$, then the value of $\tan (\theta -15^{\circ})$is equal to:
Option 1: $0$
Option 3: $\sqrt{3}$
Option 4: $\frac{1}{\sqrt{3}}$
Question : If $4 \sin ^2 \mathrm{A}-3=0$ and $0 \leq \mathrm{A} \leq 90^{\circ}$, then $3 \sin \mathrm{A}-4 \sin ^3 \mathrm{A}$ is:
Option 2: $\sqrt{\frac{3}{2}}$
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