Question : If $\cos\theta+\sin\theta=\sqrt{2}\cos\theta$, then $\cos\theta-\sin\theta$ is:
Option 1: $\sqrt{2}\tan\theta$
Option 2: $-\sqrt{2}\cos\theta$
Option 3: $-\sqrt{2}\sin\theta$
Option 4: $\sqrt{2}\sin\theta$
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Correct Answer: $\sqrt{2}\sin\theta$
Solution : Given: $\cos\theta+\sin\theta=\sqrt{2}\cos\theta$ Squaring on both sides, we get, $\cos^{2}\theta+\sin^{2}\theta+2\cos\theta\sin\theta=2\cos^{2}\theta$ $⇒\cos^{2}\theta-\sin^{2}\theta=2\cos\theta\sin\theta$ $⇒(\cos\theta+\sin\theta)(\cos\theta-\sin\theta)=2\cos\theta\sin\theta$ $⇒\sqrt{2}\cos\theta(\cos\theta-\sin\theta)=2\cos\theta\sin\theta$ $\therefore \cos\theta-\sin\theta=\sqrt{2}\sin\theta$ Hence, the correct answer is $\sqrt{2}\sin\theta$.
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Question : If $r\sin\theta=1$, $r\cos\theta=\sqrt{3}$, then the value of $(\sqrt{3}\tan\theta+1)$ is:
Option 1: $\sqrt{3}$
Option 2: $\frac{1}{\sqrt{3}}$
Option 3: $1$
Option 4: $2$
Question : If $\cos \theta+\cos ^2 \theta=1$, find the value of $\sqrt{\sin ^4 \theta+\cos ^2 \theta}$.
Option 1: $\sqrt{2} \cos \theta$
Option 2: $2 \operatorname{cos} \theta$
Option 3: $\sqrt{2} \operatorname{sin} \theta$
Option 4: $2 \operatorname{sin} \theta$
Question : If $\tan ^2 \theta+\tan ^4 \theta=1$, then:
Option 1: $\cot ^2 \theta+\cot ^4 \theta=1$
Option 2: $\cos ^2 \theta+\cos ^4 \theta=1$
Option 3: $\sin ^2 \theta+\sin ^4 \theta=1$
Option 4: $\operatorname{cosec}^2 \theta+\sec ^4 \theta=1$
Question : If $\frac{\sin ^2 \theta}{\cos ^2 \theta-3 \cos \theta+2}=1, \theta$ lies in the first quadrant, then the value of $\frac{\tan ^2 \frac{\theta}{2}+\sin ^2 \frac{\theta}{2}}{\tan \theta+\sin \theta}$ is:
Option 1: $\frac{2 \sqrt{3}}{27}$
Option 2: $\frac{5 \sqrt{3}}{27}$
Option 3: $\frac{2 \sqrt{3}}{9}$
Option 4: $\frac{7 \sqrt{3}}{54}$
Question : If $7 \sin ^2 \theta+4 \cos ^2 \theta=5$ and $\theta$ lies in the first quadrant, then what is the value of $\frac{\sqrt{3} \sec \theta+\tan \theta}{\sqrt{2} \cot \theta-\sqrt{3} \cos \theta}$?
Option 1: $2(1+\sqrt{2})$
Option 2: $3 \sqrt{2}$
Option 3: $2(\sqrt{2}-1)$
Option 4: $4 \sqrt{2}$
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