Question : If $\sin \theta=\frac{8}{17}$, where $\theta$ is an acute angle, then what is the value of $\tan \theta+\cot \theta ?$
Option 1: $\frac{217}{110}$
Option 2: $\frac{281}{190}$
Option 3: $\frac{289}{120}$
Option 4: $\frac{512}{321}$
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Correct Answer: $\frac{289}{120}$
Solution : Given, $\sin \theta=\frac{8}{17}$ We know that $\sin^2 \theta + \cos^2 \theta=1$. So, $(\frac{8}{17})^2+ \cos^2 \theta=1$ $⇒\cos^2 \theta = 1 - \frac{64}{289}$ $⇒\cos^2 \theta = \frac{225}{289}$ $⇒\cos \theta = \frac{15}{17}$ Now, $\tan \theta+\cot \theta = \frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}$ $⇒\tan \theta+\cot \theta = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta\cos \theta}$ $⇒\tan \theta+\cot \theta = \frac{1}{\frac{8}{17}\times \frac{15}{17}}$ $⇒\tan \theta+\cot \theta = \frac{289}{120}$ Hence, the correct answer is $\frac{289}{120}$.
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