MCQ
If $\frac{3\pi}{4}<\text{a}<\pi,$ then $\sqrt{2\cot\text{a}+\frac{1}{\sin^\text{a}}}$ is equal to:
- A$1-\cot\text{a}$
- B$1+\cot\text{a}$
- C$-1+\cot\text{a}$
- D$-1-\cot\text{a}$
Solution:
We have:
$\sqrt{2\cot\alpha+\frac{1}{\sin\alpha}}$
$=\sqrt{\frac{2\cos\alpha}{\sin\alpha}+\frac{1}{\sin^2\alpha}}$
$=\sqrt{\frac{2\sin\alpha\cos\alpha+1}{\sin\alpha}}$
$=\sqrt{\frac{2\sin\alpha\cos\alpha+\sin^2\alpha+\cos^2\alpha}{\sin\alpha}}$
$=\sqrt{\frac{(\sin\alpha+\cos\alpha)^2}{\sin^2\alpha}}$
$=\sqrt{(1+\cot\alpha)^2}$
$=|1+\cot\alpha|$
$=-(1+\cot\alpha) $ $$ $[\text{ when}\frac{3\pi}{4}<\alpha<\pi,\cot\alpha<-1\Rightarrow\cot\alpha+1<0\big]$
$=-1-\cot\alpha$
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The distance between the orthocentre and circumcentre of the triangle with vertices (1, 2), (2, 1) and $\Big(\frac{3+\sqrt{3}}{2},\frac{3+\sqrt{3}}{2}\Big)$ is:
$\lim\limits_{\text{x} \rightarrow0}\frac{\sin\text{x}}{\sqrt{\text{x}+1}-\sqrt{1-\text{x}}}$ is:
$2$
$0$
$1$
$-1$