Question : If $3 \tan \theta=2$, what is the value of $\frac{(3 \sin \theta-\cos \theta)}{(3 \sin \theta+\cos \theta)}?$
Option 1: $1$
Option 2: $3$
Option 3: $\sqrt3$
Option 4: $\frac{1}{3}$
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Correct Answer: $\frac{1}{3}$
Solution : $3 \tan \theta = 2$ $⇒\tan \theta = \frac{2}{3}$ Now, $\frac{(3 \sin \theta-\cos \theta)}{(3 \sin \theta+\cos \theta)}$ Dividing numerator and denominator by $\cos \theta$, we get, $=\frac{(3\tan \theta - 1)}{(3\tan \theta+1)}$ $=\frac{3 \times \frac{2}{3} - 1}{3\times \frac{2}{3}+1}$ $=\frac{1}{3}$ Hence, the correct answer is $\frac{1}{3}$.
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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 : 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 $\sin \theta-\cos \theta=0$, then what is the value of $\sin ^2 \theta+\tan ^2 \theta$ ?
Option 1: $\frac{1}{2}$
Option 2: $1$
Option 3: $\frac{4}{5}$
Option 4: $\frac{3}{2}$
Question : If $\tan\theta =\frac{3}{4}$, then the value of $\frac{4\sin^{2}\theta–2\cos^{2}\theta}{4\sin^{2}\theta+3\cos^{2}\theta}$ is equal to:
Option 1: $\frac{1}{21}$
Option 2: $\frac{2}{21}$
Option 3: $\frac{4}{21}$
Option 4: $\frac{8}{21}$
Question : If $\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}=3$, then the value of $\sin^{4}\theta$ is:
Option 1: $\frac{2}{5}$
Option 2: $\frac{1}{5}$
Option 3: $\frac{16}{25}$
Option 4: $\frac{3}{5}$
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