Question
If $\sin \theta=\frac{3}{5}$, evaluate $\frac{\operatorname{cosec} \theta-\cot \theta}{2 \cot \theta}$
✓
Answer
$\sin \theta=\frac{3}{5}$
$\cos ^2 \theta=1-\sin ^2 \theta$
$\cos ^2 \theta=1-\frac{9}{25}=\frac{16}{25}$
$\cos \theta=\frac{4}{5}$
$\cot \theta=\frac{\cos \theta}{\sin \theta}=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{3}$
$\operatorname{cosec} \theta=\frac{1}{\sin \theta}=\frac{5}{3}$
Now substituting these values in the given equation,
$\frac{\operatorname{cosec} \theta-\cot \theta}{2 \cot \theta}=\frac{\frac{5}{3}-\frac{4}{3}}{2 \cdot \frac{4}{3}}$
$\Rightarrow \frac{\frac{5}{3}-\frac{4}{3}}{2 \cdot \frac{4}{3}}=\frac{1}{3} \cdot \frac{3}{8}=\frac{1}{8}$
Hence, $\frac{\operatorname{cosec} \theta-\cot \theta}{2 \cot \theta}=\frac{1}{8}$
$\cos ^2 \theta=1-\sin ^2 \theta$
$\cos ^2 \theta=1-\frac{9}{25}=\frac{16}{25}$
$\cos \theta=\frac{4}{5}$
$\cot \theta=\frac{\cos \theta}{\sin \theta}=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{3}$
$\operatorname{cosec} \theta=\frac{1}{\sin \theta}=\frac{5}{3}$
Now substituting these values in the given equation,
$\frac{\operatorname{cosec} \theta-\cot \theta}{2 \cot \theta}=\frac{\frac{5}{3}-\frac{4}{3}}{2 \cdot \frac{4}{3}}$
$\Rightarrow \frac{\frac{5}{3}-\frac{4}{3}}{2 \cdot \frac{4}{3}}=\frac{1}{3} \cdot \frac{3}{8}=\frac{1}{8}$
Hence, $\frac{\operatorname{cosec} \theta-\cot \theta}{2 \cot \theta}=\frac{1}{8}$
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