Question : If $\tan \theta=\frac{4}{3}$, then the value of $\frac{3\sin \theta+ 2\cos \theta}{3\sin \theta – 2 \cos \theta}$ is:
Option 1: $\frac{1}{2}$
Option 2: $1\frac{1}{2}$
Option 3: $3$
Option 4: $–3$
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Correct Answer: $3$
Solution : $\frac{3\sin\theta+2\cos\theta}{3\sin\theta–2\cos\theta}$ Divide both sides by $\cos\theta$, = $\frac{\frac{3\sin\theta}{\cos\theta}+2}{\frac{3\sin\theta}{\cos\theta}–2}$ = $\frac{3\tan\theta+2}{3\tan\theta–2}$ = $\frac{3×\frac{4}{3}+2}{3×\frac{4}{3}-2}$ = $\frac{6}{2}$ = 3 Hence, the correct answer is $3$.
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Question : If $\tan\theta=1$, then the value of $\frac{8\sin\theta\:+\:5\cos\theta}{\sin^{3}\theta\:–\:2\cos^{3}\theta\:+\:7\cos\theta}$ is:
Question : Which of the following is equal to $[\frac{\tan \theta+\sec \theta–1}{\tan \theta–\sec \theta+1}]$?
Question : What is $\tan \frac{\theta}{2}$?
Question : $\frac{1+\sin \theta}{\cos \theta}$ is equal to which of the following (where $\left.\theta \neq \frac{\pi}{2}\right)?$
Question : If $\sqrt{3} \tan \theta=3 \sin \theta$, then what is the value of $\sin ^2 \theta-\cos ^2 \theta$?
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