Question
If $\sin\theta=\frac{3}{4},$ prove that $\sqrt{\frac{\text{cosec}^2\theta-\cot^2\theta}{\sec^2\theta-1}}=\frac{\sqrt{7}}{3}.$
✓
Answer
We have, $\sin\theta=\frac{3}{4}$ In $\triangle\text{ABC},$
$\text{AC}^2=\text{AB}^2+\text{BC}^2$ $\Rightarrow(4)^2=(3)^2+\text{BC}^2$ $\Rightarrow\text{BC}^2=16-9$ $\Rightarrow\text{BC}^2=2$ $\Rightarrow\text{BC}=\sqrt{7}$ $\therefore\text{cosec }\theta=\frac{4}{3},\sec\theta=\frac{4}{\sqrt{7}}$ and $\cot\theta=\frac{\sqrt{7}}{3}$ Now, $\text{L}.\text{H}.\text{S}=\sqrt{\frac{\text{cosec}^2\theta-\cot^2\theta}{\sec^2\theta-1}}$ $=\sqrt{\frac{\Big(\frac{4}{3}\Big)^2-\Big(\frac{\sqrt{7}}{3}\Big)^2}{\Big(\frac{4}{\sqrt{7}}\Big)^2-1}}$ $=\sqrt{\frac{\frac{16}{9}-\frac{7}{9}}{\frac{16}{7}-1}}$ $=\sqrt{\frac{\frac{9}{9}}{\frac{16-7}{7}}}$ $=\sqrt{\frac{7}{9}}$ $=\frac{\sqrt{7}}{3}$ $= \text{R.H.S}$
$\text{AC}^2=\text{AB}^2+\text{BC}^2$ $\Rightarrow(4)^2=(3)^2+\text{BC}^2$ $\Rightarrow\text{BC}^2=16-9$ $\Rightarrow\text{BC}^2=2$ $\Rightarrow\text{BC}=\sqrt{7}$ $\therefore\text{cosec }\theta=\frac{4}{3},\sec\theta=\frac{4}{\sqrt{7}}$ and $\cot\theta=\frac{\sqrt{7}}{3}$ Now, $\text{L}.\text{H}.\text{S}=\sqrt{\frac{\text{cosec}^2\theta-\cot^2\theta}{\sec^2\theta-1}}$ $=\sqrt{\frac{\Big(\frac{4}{3}\Big)^2-\Big(\frac{\sqrt{7}}{3}\Big)^2}{\Big(\frac{4}{\sqrt{7}}\Big)^2-1}}$ $=\sqrt{\frac{\frac{16}{9}-\frac{7}{9}}{\frac{16}{7}-1}}$ $=\sqrt{\frac{\frac{9}{9}}{\frac{16-7}{7}}}$ $=\sqrt{\frac{7}{9}}$ $=\frac{\sqrt{7}}{3}$ $= \text{R.H.S}$
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