Question : If $\tan \theta -\cot \theta =0$, find the value of $\sin \theta +\cos \theta$.
Option 1: $0$
Option 2: $1$
Option 3: $\sqrt{2}$
Option 4: $2$
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Correct Answer: $\sqrt{2}$
Solution : Given: $\tan\theta -\cot\theta =0$ $⇒\tan\theta=\cot\theta$ $\therefore\theta = 45^{\circ}$ $\therefore \sin\theta +\cos\theta = \sin45^{\circ} + \cos45^{\circ}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt2}=\sqrt2$ Hence, the correct answer is $\sqrt2$.
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Question : If $\sin\theta+\cos\theta=\sqrt{2}\cos\theta$, then the value of $\cot\theta$ is:
Option 1: $\sqrt{2}+1$
Option 2: $\sqrt{2}-1$
Option 3: $\sqrt{3}-1$
Option 4: $\sqrt{3}+1$
Question : The value of $\frac{\sin\theta-2\sin^{3}\theta}{2\cos^{3}\theta-\cos\theta}$ is equal to:
Option 1: $\sin\theta$
Option 2: $\cos\theta$
Option 3: $\tan\theta$
Option 4: $\cot\theta$
Question : If $6 \cot \theta=5$, then find the value of $\frac{(6 \cos \theta+\sin \theta)}{(6 \cos\theta-4 \sin\theta)}$
Option 1: 5
Option 2: 1
Option 3: 6
Option 4: 0
Question : If $\tan \theta+\sin \theta=A$ and $\tan \theta-\sin \theta=B$, then what is the value of $A^2-B^2$?
Option 1: $\tan \theta \sin \theta$
Option 2: $4 \cot \theta$
Option 3: $4 \tan \theta \sin \theta$
Option 4: $2 \tan \theta \sin \theta$
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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