Question
If $\text{cosec}\theta=2,$ show that $\Big(\cot\theta+\frac{\sin\theta}{1+\cos\theta}\Big)=2.$
✓
Answer
Given: $\text{cosec}\theta=\frac{\text{AC}}{\text{BC}}=\frac21$
Let BC = 1k and AC = 2k Where k is positive Let us draw a
$\triangle\text{ABC}$ in which
$\angle\text{B}=90^\circ$ and $\angle\text{A}=\theta$
By Pythagoras theorem, we have
$(AC)^2 = (AB)^2 + (BC)^2$
$\Rightarrow (AB)^2 = (AC)^2 - (BC)^2$
$=\Big[(2\text{k})^2-(1\text{k})^2\Big]=\big(4\text{k}^2-1\text{k}^2\big)=3\text{k}^2$
$\Rightarrow(\text{AB})=\sqrt{3}\text{k}$
$\sin\theta=\frac{\text{BC}}{\text{AC}}=\frac{1\text{k}}{2\text{k}}=\frac{1}{2}$
$\cos\theta=\frac{\text{AB}}{\text{AC}}=\frac{\sqrt{3}\text{k}}{2\text{k}}=\frac{\sqrt{3}}{2}$
$\cot\theta=\frac{\cos\theta}{\sin\theta}=\Big(\frac{\sqrt{3}}{2}\times\frac21\Big)=\sqrt{3}$
$\Rightarrow\Big[\cot\theta+\frac{\sin\theta}{1+\cos\theta}\Big]\Bigg[\sqrt{3}+\frac{\frac12}{1+\frac{\sqrt{3}}{2}}\Bigg]$
$=\Big(\sqrt{3}+\frac{1}{2+\sqrt{3}}\Big)=\Big(\frac{2\sqrt{3}+3+1}{2+\sqrt{3}}\Big)$
$=\Big(\frac{2\sqrt{3}+4}{2+\sqrt{3}}\Big)=2\Big(\frac{\sqrt{3}+2}{2+\sqrt{3}}\Big)=2$
Hence, $\Big[\cot\theta+\frac{\sin\theta}{1+\cos\theta}\Big]=2$
Let BC = 1k and AC = 2k Where k is positive Let us draw a
$\triangle\text{ABC}$ in which
$\angle\text{B}=90^\circ$ and $\angle\text{A}=\theta$
By Pythagoras theorem, we have
$(AC)^2 = (AB)^2 + (BC)^2$
$\Rightarrow (AB)^2 = (AC)^2 - (BC)^2$
$=\Big[(2\text{k})^2-(1\text{k})^2\Big]=\big(4\text{k}^2-1\text{k}^2\big)=3\text{k}^2$
$\Rightarrow(\text{AB})=\sqrt{3}\text{k}$
$\sin\theta=\frac{\text{BC}}{\text{AC}}=\frac{1\text{k}}{2\text{k}}=\frac{1}{2}$
$\cos\theta=\frac{\text{AB}}{\text{AC}}=\frac{\sqrt{3}\text{k}}{2\text{k}}=\frac{\sqrt{3}}{2}$
$\cot\theta=\frac{\cos\theta}{\sin\theta}=\Big(\frac{\sqrt{3}}{2}\times\frac21\Big)=\sqrt{3}$
$\Rightarrow\Big[\cot\theta+\frac{\sin\theta}{1+\cos\theta}\Big]\Bigg[\sqrt{3}+\frac{\frac12}{1+\frac{\sqrt{3}}{2}}\Bigg]$
$=\Big(\sqrt{3}+\frac{1}{2+\sqrt{3}}\Big)=\Big(\frac{2\sqrt{3}+3+1}{2+\sqrt{3}}\Big)$
$=\Big(\frac{2\sqrt{3}+4}{2+\sqrt{3}}\Big)=2\Big(\frac{\sqrt{3}+2}{2+\sqrt{3}}\Big)=2$
Hence, $\Big[\cot\theta+\frac{\sin\theta}{1+\cos\theta}\Big]=2$
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