Question : If $6 \sec \theta=10$, then find the value of $\frac{5 \operatorname{cosec} \theta-3 \cot \theta}{4 \cos \theta+3 \sin \theta}$.
Option 1: $\frac{2}{3}$
Option 2: $\frac{3}{2}$
Option 3: $\frac{5}{6}$
Option 4: $\frac{6}{5}$
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Correct Answer: $\frac{5}{6}$
Solution :
Given, $6 \sec \theta=10$
⇒ $\sec\theta=\frac{10}{6}=\frac53$
We know, $\sec\theta = \frac{1}{\cos\theta},\sin^2\theta+\cos^2\theta=1\text{ and }\operatorname{cosec\theta}=\frac{1}{\sin\theta}$
⇒ $\cos\theta=\frac{3}{5}$
⇒ $\sin\theta=\sqrt{1-\cos^2\theta}$
⇒ $\sin\theta=\sqrt{1-(\frac{3}{5})^2}$
⇒ $\sin\theta=\sqrt{1-\frac{9}{25}}$
⇒ $\sin\theta=\sqrt{\frac{25-9}{25}}$
⇒ $\sin\theta=\sqrt{\frac{16}{25}}$
⇒ $\sin\theta=\frac{4}{5}$
⇒ $\operatorname{cosec\theta}=\frac54$
⇒ $\cot\theta=\frac{\cos\theta}{\sin\theta}=\frac{3}{4}$
So, $\frac{5 \operatorname{cosec} \theta-3 \cot \theta}{4 \cos \theta+3 \sin \theta}=\frac{5\times\frac54-3\times\frac34}{4\times\frac35+3\times\frac45}=\frac{\frac{25}{4}-\frac{9}{4}}{\frac{12}{5}+\frac{12}{5}}=\frac{\frac{16}{4}}{\frac{24}{5}}=\frac{16\times5}{4\times 24}=\frac56$
Hence, the correct answer is $\frac56$.
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