Question : If $(\sin \theta-\cos \theta)=0$, then the value of $\sin\;(\pi-\theta)+\sin \left(\frac{\pi}{2}-\theta\right)$ is:
Option 1: $1$
Option 2: $0$
Option 3: $\sqrt{3}$
Option 4: $\sqrt{2}$
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Correct Answer: $\sqrt{2}$
Solution : $(\sin \theta-\cos \theta)=0$ $⇒\sin \theta=\cos \theta$ $⇒\frac{\sin \theta}{\cos \theta}=1$ $⇒\tan\theta=1=\tan45°$ $\therefore \theta = 45°$ $\sin\;(\pi-\theta)+\sin \left(\frac{\pi}{2}-\theta\right)$ $=\sin\;(180°-45°)+\sin (90°-45°)$ $= \sin135°+\cos 45°$ $=\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}$ [$\because \sin 135°=\sin(90°+45°)=\frac{1}{\sqrt{2}}$] $ = \sqrt{2}$ Hence, the correct answer is $ \sqrt{2}$.
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Question : What is the value of $\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}+\frac{\sin \theta-\cos \theta}{\sin \theta+\cos \theta}$?
Option 1: $\frac{1}{\left(\sin ^2 \theta-\cos ^2 \theta\right)}$
Option 2: $2\left(\sin ^2 \theta-\cos ^2 \theta\right)$
Option 3: $\frac{2}{\left(\sin ^2 \theta-\cos ^2 \theta\right)}$
Option 4: $\sin ^2 \theta-\cos ^2 \theta$
Question : If $\sin \theta \cos \theta=\frac{1}{\sqrt{3}}$ then the value of $\left(\sin ^4 \theta+\cos ^4 \theta\right)$ is:
Option 2: $\frac{5}{3}$
Option 3: $\frac{2}{3}$
Option 4: $\frac{1}{3}$
Question : If $\sin \theta \cos \theta=\frac{\sqrt{2}}{3}$,then the value of $\left(\sin ^6 \theta+\cos ^6 \theta\right)$ is:
Option 1: $\frac{1}{3}$
Option 2: $\frac{4}{3}$
Option 4: $\frac{5}{3}$
Question : If $\theta$ is an acute angle and $\sin \theta \cos \theta=2 \cos ^3 \theta-\frac{1}{4} \cos \theta$, then the value of $\sin \theta$ is:
Option 1: $\frac{\sqrt{15}-1}{8}$
Option 2: $\frac{\sqrt{15}-1}{4}$
Option 3: $\frac{\sqrt{15}+1}{4}$
Option 4: $\frac{\sqrt{15}-1}{2}$
Question : If $x=a\left ( \sin\theta+\cos\theta \right )$ and $y=b\left ( \sin\theta-\cos\theta \right )$, then the value of $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}$ is:
Option 1: 4
Option 2: 3
Option 3: 1
Option 4: 2
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